# Suppose that f(x)= x- 3(x^(1/3)) (A) Find all critical values of f. If there are no critical values, enter -1000. (B) Use interval notation to indicate where f(x) is increasing. (C) Use interval notation to indicate where f(x) is decreasing.(D) Find the x-coordinates of all local maxima of f. If there are no local maxima, enter -1000. (E) Find the x-coordinates of all local minima of f. If there are no local minima, enter -1000. (F) Use interval notation to indicate where f(x) is concave up.(G) Use interval notation to indicate where f(x) is concave down.(H) Find all inflection points of f. If there are no inflection points, enter -1000.

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)...

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.

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