## 170- Quiz 2

Quiz 2 is here.

Solutions follow.

Problem 1 asks for the graph of $\displaystyle f(x)=1+\frac1{x-1}$, beginning with the graph of $\displaystyle y=\frac1x.$

For this, note that the new graph is the result of translating the original graph one unit to the right (on account of $x$ being replaced with $x-1$) and one unit up (on account of $f(x)$ being 1 larger than $\displaystyle \frac1{x-1}$).

In the graph (click to enlarge), I have highlighted both axes, and drawn in blue the required curve and in black the graph of $y=1/x$, for comparison. Note that the required graph goes through the origin; this corresponds to $(-1,-1)$ being in the original graph.

Problem 2 asks to find $\displaystyle \frac{f(x+\triangle x)-f(x)}{\triangle x}$ when $f(x)=mx+b$, to simplify this as much as possible, and to determine what happens when $\triangle x$ approaches 0. In other words, we want to find a formula for $f'(x)$.

We have $f(x)=mx+b$, so $f(x+\triangle x)=m(x+\triangle x)+b=mx+m\triangle x+b$. It follows that $f(x+\triangle x)-f(x)=(mx+m\triangle x+b)-(mx+b)$ $=mx+m\triangle x+b-mx-b=m\triangle x$ and therefore

$\displaystyle \frac{f(x+\triangle x)-f(x)}{\triangle x}=\frac{m\triangle x}{\triangle x}=m.$

Note that this is a constant, and therefore its value does not change as $\triangle x$ approaches 0, i.e., $f'(x)=m$. This corresponds to the fact that straight lines have the same slope at all points.