Last calc question, I swear. This is my last question for the night.
For the given function, find the points on the graph at which the tangent line has slope 1.
y = 1/3x^3 - 1/2x^2 +x
If you mean $\displaystyle y=\frac{1}{3x^3}-\frac{1}{2x^2}+x$, the derivative would then be $\displaystyle y^{/}=-\frac{1}{x^4}+\frac{1}{x^3}+1$. To see where it's equal to 1:
$\displaystyle 1=-\frac{1}{x^4}+\frac{1}{x^3}+1$
Multiply through by $\displaystyle x^4$ to clean this up:
$\displaystyle x^4=-1+x+x^4\rightarrow x=1$. Is this what you're looking for?