I need to find the absolute max and min values of f(x)=2x-3x^2/3 on the closed interval [-1,8]PLEASE HELP!!

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rcmath | High School Teacher | (Level 1) Associate Educator

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Let's start by finding the first and second derivative.


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


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

`f''(x)=2/3*x^(-4/3) =>`


Notice that f'' is positive for all values of x. Which means of an inflection point exit, it will be a minimum.

Let's solve f'(x)=0 to find the relative minimum.

` ` `2-2/root(3)(x)=0 => 2/root(3)(x)=2 =>root(3)(x)=1=>x=1`


So point (1,-1) is a relative min, to check if it is an absolute we need to find the value of the function at the end points.

`f'(-1)=2(-1)-3(-1)^(2/3)=> f(-1)=-2-3=-5`


Thus the absolute min is at the end point (-1,-5), and the absolute max is at the other endpoint (8,4).

The following graph confirm our findings.



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