What are the local extrema, if any, of #f (x) = 2x^4-36x^2+5 #?

Answer 1

#x={-3,0,3}#

Local extrema occur whenever the slope is equal to 0 so we must first find the derivative of the function, set it equal to 0, and then solve for x to find all x's for which there are local extrema.

Using the power-down rule we can find that #f'(x)=8x^3-72x#. Now set it equal to 0. #8x^3-72x=0#. To solve, factor out an #8x# to get #8x(x^2-9)=0# then using the rule of the difference of two squares split #x^2-9# into its two factors to get #8x(x+3)(x-3)=0#. Now set each of these separately equal to 0 because the entire expression will be 0 when any of the terms are 0.
This gives you 3 equations: #8x=0#, #x+3=0#, and #x-3=0#. To solve the first one divide both sides by 8 to get #x=0#. For the second, subtract 3 from both sides to get #x=-3#. Lastly, for the third, add 3 to both sides to get #x=3#. These are all the x-values where local extrema will occur. Hope I helped!
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Answer 2

To find the local extrema of ( f(x) = 2x^4 - 36x^2 + 5 ), we first take the derivative of the function and set it equal to zero to find critical points. Then we use the second derivative test to determine whether these critical points correspond to local maxima, minima, or points of inflection.

  1. Find the derivative of ( f(x) ): [ f'(x) = 8x^3 - 72x ]

  2. Set ( f'(x) = 0 ) and solve for ( x ) to find critical points: [ 8x^3 - 72x = 0 ] [ 8x(x^2 - 9) = 0 ] [ x(x - 3)(x + 3) = 0 ]

    Critical points: ( x = 0, x = 3, x = -3 )

  3. Find the second derivative of ( f(x) ): [ f''(x) = 24x^2 - 72 ]

  4. Evaluate ( f''(x) ) at each critical point:

    • For ( x = 0 ): [ f''(0) = -72 ] Since ( f''(0) < 0 ), ( x = 0 ) corresponds to a local maximum.

    • For ( x = 3 ): [ f''(3) = 72 ] Since ( f''(3) > 0 ), ( x = 3 ) corresponds to a local minimum.

    • For ( x = -3 ): [ f''(-3) = 72 ] Since ( f''(-3) > 0 ), ( x = -3 ) corresponds to a local minimum.

Therefore, the local extrema of the function are:

  • Local maximum at ( x = 0 )
  • Local minimums 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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