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Section 3.4 : The Definition of a Function

13. The difference quotient for the 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 the function $$f\left( x \right) = 2{x^2} - x$$.

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We’ll work this problem in parts. First let’s compute $$f\left( {x + h} \right)$$.

$f\left( {x + h} \right) = 2{\left( {x + h} \right)^2} - \left( {x + h} \right) = 2\left( {{x^2} + 2xh + {h^2}} \right) - \left( {x + h} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{2{x^2} + 4xh + 2{h^2} - x - h}}$ Show Step 2

Now we’ll compute $$f\left( {x + h} \right) - f\left( x \right)$$ and do a little simplification.

\begin{align*}f\left( {x + h} \right) - f\left( x \right) & = 2{x^2} + 4xh + 2{h^2} - x - h - \left( {2{x^2} - x} \right)\\ & = 2{x^2} + 4xh + 2{h^2} - x - h - 2{x^2} + x = \require{bbox} \bbox[2pt,border:1px solid black]{{4xh + 2{h^2} - h}}\end{align*}

Be careful with the parenthesis when subtracting $$f\left( x \right)$$. We need to subtract the function and so we need parenthesis around the whole thing to make sure we do subtract the function.

Show Step 3

We can now finish the problem by computing the full difference quotient.

$\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{{4xh + 2{h^2} - h}}{h} = \frac{{h\left( {4x + 2h - 1} \right)}}{h} = \require{bbox} \bbox[2pt,border:1px solid black]{{4x + 2h - 1}}$