A) You need to solve the equation f'(x)=0 to check if the function has critical values such that:

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

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

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

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

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

You need to solve f'(x)...

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A) You need to solve the equation f'(x)=0 to check if the function has critical values such that:

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

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

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

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

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

You need to solve f'(x) = 0 such that:

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

`(root(3)x^2) = 1 =gt x^2 = 1^3 =gt x^2 = 1 =gt x = +-1`

**Hence, the critical values of the function are x = -1 and x = 1.**

B)**The derivative is positive for `x in (-oo,-1)U(1,oo)` , hence the function increases over `(-oo,-1)U(1,oo)` .**

C) **The derivative is negative for `x in (-1,1),` hence the function decreases over (-1,1).**

D) **The function reaches its maximum at x = -1**.

E) **The function reaches its minimum at x = 1**.

F)You need to find f''(x) to find where the function is concave up or concave down.

`f''(x) = ((2/3)(x^(2/3 - 1)))/(x^(4/3))`

`f''(x) = 2/(3xroot(3)(x^2))`

**Notice that the second derivative is positive for `x in R` set, hence the function is concave up over R set.**