How do you find the extrema for #f(x)=x^4-18x^2+7#?

Answer 1

This function has 3 extrema:

  • A maximum at 0 #f(0)=7#
  • 2 minima at -3 and 3 #f(-3)=f(3)=-74#
To calculete the extrema of a function you have to find points, where #f'(x)=0# first. In this case you get: #4x^3-36x=0# #4x(x^2-9)=0# #x=0 vv x=-3 xx x=3#
Now you have to check how #f'(x)# looks like in the surrounding of the points calculated above. To check the behaviour you can either draw a graph or calculate #f''(x)#
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Answer 2

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.

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

  2. 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 ).

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

  2. 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 ]

  3. 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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Answer 3

To find the extrema for the function ( f(x) = x^4 - 18x^2 + 7 ), follow these steps:

  1. Calculate the derivative of the function ( f'(x) ).
  2. Set ( f'(x) = 0 ) to find critical points.
  3. 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:

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

  2. 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 ).

  1. 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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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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