Calculus I - Functions

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Section 1.1 : Review : Functions

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8. 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 \(\displaystyle y\left( z \right) = \frac{1}{{z + 2}}\).

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Start Solution

\[y\left( {z + h} \right) = \frac{1}{{z + h + 2}}\] Show Step 2

\[y\left( {z + h} \right) - y\left( z \right) = \frac{1}{{z + h + 2}} - \frac{1}{{z + 2}} = \frac{{z + 2 - \left( {z + h + 2} \right)}}{{\left( {z + h + 2} \right)\left( {z + 2} \right)}} = \frac{{ - h}}{{\left( {z + h + 2} \right)\left( {z + 2} \right)}}\]

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.

Show Step 3

\[\frac{{y\left( {z + h} \right) - y\left( z \right)}}{h} = \frac{1}{h}\left( {y\left( {z + h} \right) - y\left( z \right)} \right) = \frac{1}{h}\left( {\frac{{ - h}}{{\left( {z + h + 2} \right)\left( {z + 2} \right)}}} \right) = \frac{{ - 1}}{{\left( {z + h + 2} \right)\left( {z + 2} \right)}}\]

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,

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