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### Section 1-1 : Review : Functions

9. The difference quotient of a function $$f\left( x \right)$$ is defined to be,

$\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

compute the difference quotient for $$\displaystyle A\left( t \right) = \frac{{2t}}{{3 - t}}$$.

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Hint : Don’t get excited about the fact that the function is now named $$A\left( t \right)$$, the difference quotient still works in the same manner it just has $$A$$’s and $$t$$’s instead of $$f$$’s and $$x$$’s now. So, compute $$A\left( {t + h} \right)$$, then compute the numerator and finally compute the difference quotient.
Start Solution
$A\left( {t + h} \right) = \frac{{2\left( {t + h} \right)}}{{3 - \left( {t + h} \right)}} = \frac{{2t + 2h}}{{3 - t - h}}$ Show Step 2
\begin{align*}A\left( {t + h} \right) - A\left( t \right) & = \frac{{2t + 2h}}{{3 - t - h}} - \frac{{2t}}{{3 - t}}\\ & = \frac{{\left( {2t + 2h} \right)\left( {3 - t} \right) - 2t\left( {3 - t - h} \right)}}{{\left( {3 - t - h} \right)\left( {3 - t} \right)}}\\ & = \frac{{6t - 2{t^2} + 6h - 2ht - \left( {6t - 2{t^2} - 2th} \right)}}{{\left( {3 - t - h} \right)\left( {3 - t} \right)}}\\ & = \frac{{6h}}{{\left( {3 - t - h} \right)\left( {3 - t} \right)}}\end{align*}

Note that, when dealing with difference quotients, it will almost always be advisable to combine rational expressions into a single term in preparation of the next step. Also, when doing this don’t forget to simplify the numerator as much as possible. With most difference quotients you’ll see a lot of cancelation as we did here.

Show Step 3
$\frac{{A\left( {t + h} \right) - A\left( t \right)}}{h} = \frac{1}{h}\left( {A\left( {t + h} \right) - A\left( t \right)} \right) = \frac{1}{h}\left( {\frac{{6h}}{{\left( {3 - t - h} \right)\left( {3 - t} \right)}}} \right) = \frac{6}{{\left( {3 - t - h} \right)\left( {3 - t} \right)}}$

In this step we rewrote the difference quotient a little to make the numerator a little easier to deal with. All that we’re doing here is using the fact that,

$\frac{a}{b} = \left( a \right)\left( {\frac{1}{b}} \right) = \left( {\frac{1}{b}} \right)\left( a \right)$