How do you find the exact relative maximum and minimum of the polynomial function of #f(x)=x^3-3x+6#?

Answer 1

Relative maximum: #8# (at #-1#) and relative minimum: #4# (at #1#)

#f(x) = x^3-3x+6#
Find critical numbers for #f# .
#f'(x) = 3x^2-3# never fails to exist and is #0# at the solutions of
#3x^2-3=0#.
The solution are #+-1#, so the critical numbers are #+-1#.

Test the critical numbers

Use either the first or second derivative test to see that

#f(-1)# is a relative maximum.
(Second derivative test: #f''(x) = 6x#, so #f''(-1) < 0# and we conclude #f(-10# is a relative maximum.)
And #f(1)# is a relative minimum.
(f''(1) > 0#, so #f(1) is a relative minimum.)

Determine the lowest and highest numbers.

Rel max: #f(-1)=(-1)^3-3(-1)+6 = 8#
Rel min: #f(1) = (1)^3-3(1)+6 = 4#
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Answer 2

To find the exact relative maximum and minimum of the polynomial function ( f(x) = x^3 - 3x + 6 ), you can follow these steps:

  1. Find the critical points by setting the derivative of ( f(x) ) equal to zero and solving for ( x ).
  2. Determine the nature of these critical points (whether they correspond to relative maximum, minimum, or inflection points) using the second derivative test.
  3. Evaluate ( f(x) ) at the critical points to find the corresponding function values, which will give you the relative maximum and minimum.

Let's proceed with the calculations:

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

Setting ( f'(x) = 0 ): [ 3x^2 - 3 = 0 ] [ x^2 - 1 = 0 ] [ (x - 1)(x + 1) = 0 ]

So, the critical points are ( x = 1 ) and ( x = -1 ).

  1. Find the second derivative of ( f(x) ): [ f''(x) = 6x ]

  2. Evaluate ( f''(x) ) at the critical points: [ f''(1) = 6(1) = 6 > 0 ] [ f''(-1) = 6(-1) = -6 < 0 ]

By the second derivative test:

  • At ( x = 1 ), since ( f''(1) > 0 ), ( f(x) ) has a relative minimum at ( x = 1 ).
  • At ( x = -1 ), since ( f''(-1) < 0 ), ( f(x) ) has a relative maximum at ( x = -1 ).

Now, evaluate ( f(x) ) at these critical points: [ f(1) = (1)^3 - 3(1) + 6 = 1 - 3 + 6 = 4 ] [ f(-1) = (-1)^3 - 3(-1) + 6 = -1 + 3 + 6 = 8 ]

So, the relative minimum is at ( (1, 4) ) and the relative maximum is at ( (-1, 8) ).

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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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