I've found hard to answer this problem, I would apreciate your help:
Find the points on the graph of f(x) = 1/3x^3+x^2-x-1 at which the slope is (a) -1, (b) 2, and (c) 0
for example for $\displaystyle f'(x) = 0$ the solution is like this:
$\displaystyle x^2+2x-1=0$
$\displaystyle X^2+2x+1-2=0$
$\displaystyle (x+1)^2-2=0$
$\displaystyle (x+1)^2=2$
Then answers are:
$\displaystyle x = -1+\sqrt2$
or
$\displaystyle x= -1-\sqrt2$