find absolute extrema question math?

Absolute max and min are respectively the greatest and least value of the function on the interval being considered. These will occur at either the endpoints or at critical points within the interval.

➊ The endpoints are x=-1 & x=2.

So, x=-1 & x=2 are possibilities for absolute extrema.

➋ Critical Points occur where the derivative is 0 or undefined.

f'(x)=2/x^⅓ - 2 is undefined for x=0 and

f'(x)=2/x^⅓ - 2 = 0 for x=1

So, x=0 & x=1 are possibilities for absolute extrema.

The function value for each of these x's from ➊ & ➋ is:

  i. f(-1) = 3(-1)^(2/3) - 2(-1) = 3 + 2 = 5                ← Absolute maximum of 5 at x= -1

 ii.  f(2) = 3(2)^(2/3) - 2(2) = 3∛(4) - 4 ≈ 0.7622

iii.  f(0) = 3(0)^(2/3) - 2(0) = 0                              ← Absolute minimum of 0 at x=0

iv.  f(1) = 3(1)^(2/3) - 2(1) = 3 - 2 = 1

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Graph -- Give it a little time to produce the curve

http://www.quickmath.com/webMathematica3/quickmath...

http://tutorial.math.lamar.edu/Classes/CalcI/AbsEx...

http://tutorial.math.lamar.edu/Classes/CalcI/Criti...

Have a good one!

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f' (x) = (2/3)(3)(x^-1/3) - 2

f'(x) = 2/x^1/3 - 2

0 = 2/(x^1/3) - 2

2 = 2/(x^1/3)

x = 1

f'(0.1) = +

f'(1.1) = -

We cant use 0 since 2x^1/3 would be undefined, and we cant use 2 because thats an endpoint.

So we established we have a local max at x = 1.

Now lets take limits as x -->∞ and -∞

x -->∞ = -∞

x --->-∞ = +∞

So sadly we have no absolute extrema : (

EDIT: ray is correct, i forgot that we were working on a closed interval. So test the endpoints and whatver has the hgihest or least values those are your absolute extrema

the by-product is 3x^2 - 6x 3x(x-2) x = 0 and a pair of in case you may desire to comprehend in the event that they are maxs or minutes, purely plug numbers on the two area of them into the by-product, otherwise, those 2 solutions are actually the extrema