How do you find the local extrema for #f(x)= x^4-4x³#?

Answer 1

There is a minimum for #x=3#.

First of all we search the zeros of the derivative.

#f'(x)=d/dx(x^4-4x^3)=4x^3-12x^2#
#f'(x)=0->4x^3-12x^2=0#
One solution is of course #x=0# the second is obtained dividing by #x^2#
#4x-12=0->x=3#.
To see which kind of points are #x=0# and #x=3# we can study the second derivative
#f''(x)=12x^2-24x#

and we have

#f''(0)=0#, #f(3)=36#.
Then when #x=0# the point is an horizontal flex (otherwise known as an inflection point), while when #x=3# it is a minimum.

We can see this also from the plot.

graph{x^4-4x^3 [-52.9, 61.54, -29.3, 27.9]}

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Answer 2

To find the local extrema for ( f(x) = x^4 - 4x^3 ), we first find the critical points by setting the derivative equal to zero and then classify them using the second derivative test.

[ f'(x) = 4x^3 - 12x^2 ]

Setting ( f'(x) = 0 ), we solve for critical points:

[ 4x^3 - 12x^2 = 0 ]

[ 4x^2(x - 3) = 0 ]

So, either ( x = 0 ) or ( x = 3 ).

To classify the critical points, we find the second derivative ( f''(x) ):

[ f''(x) = 12x^2 - 24x ]

Evaluate ( f''(x) ) at the critical points:

[ f''(0) = 12(0)^2 - 24(0) = 0 ] [ f''(3) = 12(3)^2 - 24(3) = 36(3 - 2) = 36 ]

At ( x = 0 ), the second derivative test is inconclusive. At ( x = 3 ), since ( f''(3) > 0 ), it's a local minimum.

Thus, ( f(x) = x^4 - 4x^3 ) has a local minimum at ( x = 3 ).

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Answer 3

To find the local extrema for ( f(x) = x^4 - 4x^3 ), you first find the critical points by taking the derivative of the function and setting it equal to zero. Then, you determine whether these critical points are local minima or local maxima by analyzing the sign of the second derivative at those points.

  1. Find the derivative of ( f(x) ): ( f'(x) = 4x^3 - 12x^2 ).
  2. Set ( f'(x) ) equal to zero and solve for ( x ) to find critical points: ( 4x^3 - 12x^2 = 0 ).
  3. Factor out ( 4x^2 ): ( 4x^2(x - 3) = 0 ).
  4. Set each factor equal to zero and solve for ( x ): ( 4x^2 = 0 ) gives ( x = 0 ), ( x - 3 = 0 ) gives ( x = 3 ).
  5. These critical points are ( x = 0 ) and ( x = 3 ).
  6. Find the second derivative of ( f(x) ): ( f''(x) = 12x^2 - 24x ).
  7. Evaluate ( f''(x) ) at each critical point: At ( x = 0 ), ( f''(0) = 0 - 0 = 0 ). At ( x = 3 ), ( f''(3) = 12(3)^2 - 24(3) = 108 - 72 = 36 ).
  8. Analyze the signs of ( f''(x) ) at the critical points:
    • At ( x = 0 ), ( f''(x) = 0 ), which means the test is inconclusive.
    • At ( x = 3 ), ( f''(x) = 36 ), which means the function has a local minimum at ( x = 3 ).

Therefore, the local minimum of ( f(x) ) is at ( x = 3 ). There are no local extrema at ( x = 0 ).

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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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