<s>[INST] Finding the points on the graph of the function $\frac{1}{3}x^3 + x^2 - x - 1$ at which the slope is $-1$; $2$ Please help me. Find the points on the graph of $\frac{1}{3}x^3 + x^2 - x - 1$ at which the slope is 
(a) $-1$; (b) $2$. 
I don't know where to begin.
 [/INST] The slope of the graph is equal to the derivative $$\frac{d}{dx}\left(\frac13x^3+x^2-x-1\right)=x^2+2x-1$$ Now solve for (a) $$x^2+2x-1=-1 \iff x^2+2x=0 \iff x(x+2)=0$$ and for (b) $$x^2+2x-1=2 \iff x^2+2x-3=0 \iff (x-1)(x+3)=0$$
</s>