How do use the first derivative test to determine the local extrema # f(x) = 3x^48x^390x^2+50#?
Find the critical numbers and test for min/max using the first derivative test.
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To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 \To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):
To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):
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Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

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Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

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Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zeroTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find criticalTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solveTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points,To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve forTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solveTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for (To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve (To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( xTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x \To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x )To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) toTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to findTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24xTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the criticalTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.
To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2 To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.
3To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.
3.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

DetermineTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180xTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x =To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the signTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign ofTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of (To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 \To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( fTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 )To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) forTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(xTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for (To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( xTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.
To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x \To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.
4To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ). To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.
4.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ). 4To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

IfTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ). 4.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If (To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

AfterTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( fTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After findingTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(xTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the criticalTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x)To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical pointsTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) \To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points,To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) )To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluateTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changesTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes signTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the signTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign fromTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign ofTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive toTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the firstTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at aTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivativeTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical pointTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point,To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervalsTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, thenTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formedTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed byTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the functionTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has aTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical pointsTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a localTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points. To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximumTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points. 5To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum atTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points. 5.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at thatTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

IfTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. IfTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the signTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If (To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positiveTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( fTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative atTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at aTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(xTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical pointTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) \To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, itTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changesTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates aTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign fromTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a localTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negativeTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximumTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative toTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. IfTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positiveTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If itTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive atTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changesTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at aTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes fromTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a criticalTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negativeTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical pointTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative toTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point,To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positiveTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, thenTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive,To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, itTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the functionTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicatesTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function hasTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates aTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has aTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a localTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a localTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum. To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimumTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum. 6To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at thatTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum. 6.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that pointTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

IfTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.
To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.
5To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the signTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.
5.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign doesTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

IfTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does notTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If (To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not changeTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( fTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change atTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at aTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x)To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical pointTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) \To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point,To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) doesTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or ifTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

OnceTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if thereTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once youTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no criticalTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identifiedTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the testTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the criticalTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test isTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the typeTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusiveTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive forTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extremaTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for thatTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema atTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that intervalTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at eachTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval. To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each pointTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval. 7To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point,To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval. 7.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, youTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

FinallyTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you canTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally,To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can stateTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verifyTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the localTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the pointsTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the local extTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the points foundTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the local extremaTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the points found byTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the local extrema ofTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the points found by theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the local extrema of theTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the points found by the firstTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the local extrema of the functionTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the points found by the first derivativeTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the local extrema of the function.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the points found by the first derivative testTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the local extrema of the function.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the points found by the first derivative test by plTo use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function ( f(x) ), denoted as ( f'(x) ).

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points.

Determine the sign of ( f'(x) ) in the intervals defined by the critical points. Use test points within each interval to determine the sign.

If ( f'(x) ) changes sign from positive to negative at a critical point, then the function has a local maximum at that point. If ( f'(x) ) changes sign from negative to positive at a critical point, then the function has a local minimum at that point.

If ( f'(x) ) does not change sign at a critical point, then the test is inconclusive for that point.

Once you've identified the critical points and determined the type of extrema at each point, you can state the local extrema of the function.To use the first derivative test to determine the local extrema of ( f(x) = 3x^4  8x^3  90x^2 + 50 ):

Find the first derivative of the function: ( f'(x) = 12x^3  24x^2  180x ).

Set ( f'(x) = 0 ) and solve for critical points.

To find critical points, solve ( 12x^3  24x^2  180x = 0 ) for ( x ).

After finding the critical points, evaluate the sign of the first derivative in the intervals formed by the critical points.

If the sign changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, it indicates a local minimum.

If the sign does not change at a critical point, or if there is no critical point in an interval, the test is inconclusive for that interval.

Finally, verify the points found by the first derivative test by plugging them back into the original function to confirm they correspond to local extrema.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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