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

`f'(x) = ((9x^2+3)')/(9x^2+3)`

`f'(x) = (18x)/(9x^2+3)`

You need to solve the equation f'(x) = 0, hence since `(9x^2+3)!=0` , then `18x = 0 =gt x = 0.`

Hence, the critical value...

## See

This Answer NowStart your **subscription** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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

`f'(x) = ((9x^2+3)')/(9x^2+3)`

`f'(x) = (18x)/(9x^2+3)`

You need to solve the equation f'(x) = 0, hence since `(9x^2+3)!=0` , then `18x = 0 =gt x = 0.`

Hence, the critical value of the function is x=0.

B) The function is increasing over `(0,oo).`

C) The function only increases over `(0,oo).`

D) You need to remember that the x coordinate of local maximum is the critical value of function, hence x = 0.

E) Since the function has no minimum, then there is no x coordinate for local minimum.

F) Since the function increases over`(0,oo),` hence the function is concave up over `(0,oo).`

G) Since the function only increases over `(0,oo)` ,hence the function is not concave down over any interval.