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.


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