What are the local extrema, if any, of #f(x)= 120x^5 - 200x^3#?

Answer 1

Local maximum of #80# (at #x=-1#) and local minimum of #-80# (at #x=1#.

#f(x)= 120x^5 - 200x^3#
#f'(x)= 600x^4 - 600x^2 = 600x^2(x^2 - 1)#
Critical numbers are: #-1#, #0#, and #1#
The sign of #f'# changes from + to - as we pass #x=-1#, so #f(-1) = 80# is a local maximum.
(Since #f# is odd, we can immediately conclude that #f(1)=-80# is a relative minimum and #f(0)# is not a local extremum.)
The sign of #f'# does not change as we pass #x= 0#, so #f(0)# is not a local extremum.
The sign of #f'# changes from - to + as we pass #x=1#, so #f(1) = -80# is a local minimum.
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Answer 2

To find the local extrema of ( f(x) = 120x^5 - 200x^3 ), we first need to find its critical points. Critical points occur where the derivative of the function is zero or undefined.

  1. Find the derivative of the function: [ f'(x) = 600x^4 - 600x^2 ]

  2. Set the derivative equal to zero and solve for ( x ): [ 600x^4 - 600x^2 = 0 ] [ 600x^2(x^2 - 1) = 0 ] [ x = 0, \pm 1 ]

These are the critical points.

  1. Determine the nature of the critical points by using the first derivative test:
  • When ( x = 0 ): Substitute ( x = 0 ) into the second derivative ( f''(x) ). [ f''(0) = 0 ]

  • When ( x = 1 ): Substitute ( x = 1 ) into the second derivative ( f''(x) ). [ f''(1) = 1200 ]

  • When ( x = -1 ): Substitute ( x = -1 ) into the second derivative ( f''(x) ). [ f''(-1) = 1200 ]

Since ( f''(1) ) and ( f''(-1) ) are both positive, and ( f''(0) = 0 ), there are local minima at ( x = -1 ) and ( x = 1 ), and ( x = 0 ) is a point of inflection.

So, the local extrema of ( f(x) = 120x^5 - 200x^3 ) are:

  • Local minimum at ( x = -1 )
  • Local minimum at ( x = 1 )
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Answer 3

To find the local extrema of ( f(x) = 120x^5 - 200x^3 ):

  1. Find the derivative of the function. [ f'(x) = 600x^4 - 600x^2 ]

  2. Set the derivative equal to zero and solve for ( x ) to find critical points. [ 600x^4 - 600x^2 = 0 ] [ 600x^2(x^2 - 1) = 0 ] [ x = 0, \pm 1 ]

  3. Test the critical points in the second derivative to determine the nature of the extrema. [ f''(x) = 2400x^3 - 1200x ]

For ( x = 0 ): [ f''(0) = 0 ]

For ( x = 1 ): [ f''(1) = 2400(1)^3 - 1200(1) = 2400 - 1200 = 1200 > 0 ]

For ( x = -1 ): [ f''(-1) = 2400(-1)^3 - 1200(-1) = -2400 - (-1200) = -1200 < 0 ]

Therefore, at ( x = 1 ), there is a local minimum, and at ( x = -1 ), there is a local maximum.

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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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