How do you find the extrema for #f(x)=x^418x^2+7#?
This function has 3 extrema:
 A maximum at 0
#f(0)=7#  2 minima at 3 and 3
#f(3)=f(3)=74#
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To find the extrema of the function ( f(x) = x^4  18x^2 + 7 ), we first take the derivative of the function and set it equal to zero to find the critical points. Then, we determine whether each critical point corresponds to a relative maximum or minimum by analyzing the sign of the second derivative.

Find the first derivative: [ f'(x) = 4x^3  36x ]

Set the first derivative equal to zero and solve for ( x ) to find critical points: [ 4x^3  36x = 0 ] [ x(4x^2  36) = 0 ] [ x(x^2  9) = 0 ] [ x(x  3)(x + 3) = 0 ]
So, the critical points are ( x = 3, 0, 3 ).

Find the second derivative: [ f''(x) = 12x^2  36 ]

Evaluate the second derivative at each critical point: [ f''(3) = 12(3)^2  36 = 108  36 = 72 > 0 ] [ f''(0) = 12(0)^2  36 = 36 < 0 ] [ f''(3) = 12(3)^2  36 = 108  36 = 72 > 0 ]

Analyze the sign of the second derivative to determine the nature of the critical points:
 ( x = 3 ): ( f''(3) > 0 ), so it corresponds to a local minimum.
 ( x = 0 ): ( f''(0) < 0 ), so it corresponds to a local maximum.
 ( x = 3 ): ( f''(3) > 0 ), so it corresponds to a local minimum.
Therefore, the extrema of ( f(x) = x^4  18x^2 + 7 ) are:
 Local maximum at ( x = 0 ), ( f(0) = 7 )
 Local minima at ( x = 3 ) and ( x = 3 ), ( f(3) = 52 ) and ( f(3) = 52 ) respectively.
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To find the extrema for the function ( f(x) = x^4  18x^2 + 7 ), follow these steps:
 Calculate the derivative of the function ( f'(x) ).
 Set ( f'(x) = 0 ) to find critical points.
 Use the second derivative test or the behavior of the first derivative around critical points to determine whether each critical point corresponds to a local minimum, local maximum, or neither.
Let's begin:

Calculate the derivative of the function: [ f'(x) = 4x^3  36x ]

Set ( f'(x) = 0 ) and solve for ( x ) to find critical points: [ 4x^3  36x = 0 ] [ 4x(x^2  9) = 0 ] [ 4x(x  3)(x + 3) = 0 ]
This equation has three critical points: ( x = 0 ), ( x = 3 ), and ( x = 3 ).
 Use the second derivative test to determine the nature of each critical point: [ f''(x) = 12x^2  36 ]
For ( x = 0 ): [ f''(0) = 36 ] Since the second derivative is negative, ( x = 0 ) corresponds to a local maximum.
For ( x = 3 ): [ f''(3) = 108  36 = 72 ] Since the second derivative is positive, ( x = 3 ) corresponds to a local minimum.
For ( x = 3 ): [ f''(3) = 108  36 = 72 ] Since the second derivative is positive, ( x = 3 ) corresponds to a local minimum.
So, the function has local maximum at ( x = 0 ) and local minima at ( x = 3 ) and ( x = 3 ).
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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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