Find the range of $x$ values for which the graph is decreasing for $f(x)=3x^2-2x^3$.

The region for which the function is decreasing means that the slope has to be negative.

A function is defined on $\mathbb{R} \backslash \{1\}$ by $f(x)=\frac{x}{1-x}$. Show that $f$ is an increasing function on $\mathbb{R} \backslash \{1\}$.

We need to show that for all values in $\mathbb{R} \backslash \{1\}$, the slopes of the tangents are positive. First, differentiate the function $f$.

$f'(x)$ is positive for all values within the range. $(1-x)^2$ is a positive number since anything squared is positive. $\frac{1}{(1-x)^2}$ is positive since a positive divided by a positive, is also positive.