On what interval(s) is the function f (x) = 3x^(1/3) - x increasing

On what interval(s) is the function f (x) = 3x^(1/3) - x increasing?
a. (-1,0) U (0,1)
b. (negative infinity, -1) U(1, infinity)
c. (negative infinity, infinity)
d. F is never increasing



Suppose lim (x----> 6 - ) f(x)= infinity. What conclusion can be made about the graph of y = f (x)?
a. There is a horizontal asymptote at y= - 6
b. There is a horizontal asymptote at y=6
c. there is a vertical asymptote at x = - 6
d. there is a vertical asymptote at x=6

For the first question, you want to know when the slope of the functions changes from postive to negative. To find these points, take the first derivative (to get slope) and set to zero.
f'(x) = 3*1/3*x^(-2/3) - 1; zeros are 1 = x^-2/3 or x +/- 1.
Now calculate values of f(x) at x = -1, x=0 and x= 1 to get f(-1) = -4, f(0)=0 and f(1)=2. Then calculate values to the to the left of -1, f(-8) is = 2, and right of 1 say f(8) = -2 (-8 and +8 are nice cubics)

So the correct answer is A

Second question: The function would have a vertical asymptote at x = 6. It may be one or two-sided depending on the lim as x -> 6+. So the answer is D

Hope that helps!