How do you find the local extrema for #y=4x^3 + 7#?
No calculus answer
Using the derivative
(There are no other critical numbers to consider, so there is no extremum.)
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To find the local extrema of ( y = 4x^3 + 7 ), you first need to find its critical points. Critical points occur where the derivative of the function is either zero or undefined.

Find the derivative of the function ( y ) with respect to ( x ), denoted as ( y' ). [ y' = 12x^2 ]

Set the derivative equal to zero and solve for ( x ) to find critical points. [ 12x^2 = 0 ] [ x^2 = 0 ] [ x = 0 ]

Once you have the critical points, you can determine the nature of the extrema by using the second derivative test.
a. Find the second derivative of ( y ), denoted as ( y'' ). [ y'' = 24x ]
b. Plug the critical point(s) ( x ) into the second derivative. [ y''(0) = 24(0) = 0 ]

Analyze the sign of the second derivative at the critical point(s).
a. If ( y'' ) is positive, the function has a local minimum at that point.
b. If ( y'' ) is negative, the function has a local maximum at that point.
c. If ( y'' ) is zero or undefined, the test is inconclusive.
Since ( y''(0) = 0 ), the second derivative test is inconclusive. To determine whether ( x = 0 ) corresponds to a local minimum, maximum, or neither, you can either analyze the function around the critical point or use other methods like the first derivative test.
However, in this case, since ( y = 4x^3 + 7 ) is a cubic function, it does not have a local maximum or minimum as it either increases or decreases indefinitely. Thus, there are no local extrema for the function ( y = 4x^3 + 7 ).
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To find the local extrema for the function ( y = 4x^3 + 7 ), we first find its critical points by setting the derivative ( y' ) equal to zero and then determine the nature of these points using the second derivative test.

Find the derivative of ( y ) with respect to ( x ): [ y' = 12x^2 ]

Set ( y' ) equal to zero and solve for ( x ) to find the critical points: [ 12x^2 = 0 ] [ x = 0 ]
So, ( x = 0 ) is the only critical point of the function.

Next, find the second derivative of ( y ) with respect to ( x ): [ y'' = 24x ]

Evaluate ( y'' ) at the critical point ( x = 0 ): [ y''(0) = 24(0) = 0 ]
Since the second derivative at the critical point is zero, the second derivative test is inconclusive for determining the nature of the extremum at ( x = 0 ). Therefore, we need to analyze the function near ( x = 0 ) to determine if it has a local extremum there.
For this, we can look at the sign of the first derivative ( y' ) in intervals around ( x = 0 ):
 When ( x < 0 ), ( y' < 0 ), indicating a local maximum.
 When ( x > 0 ), ( y' > 0 ), indicating a local minimum.
So, the function ( y = 4x^3 + 7 ) has a local maximum at ( x = 0 ).
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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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