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!!

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

Posted on

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

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

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

and

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

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

`f''(x)=2/[3(root(3)(x))^4]`

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`

`f(1)=2-3=-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`

`f'(8)=2(8)-3(8)^(2/3)=16-3(2)^2=16-3*4=4`

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.