audio read-through More on Difference Quotients

Difference quotients were introduced in an earlier lesson:  Prerequisites: Function Review; Difference Quotients.

Evaluating difference quotients requires a high degree of comfort working with functions and function notation, and this earlier lesson offers a thorough review.

Then, this current lesson gives practice with difference quotient problems involving reciprocal functions and square roots.

DIFFERENCE QUOTIENTS

Let $\,f\,$ be a function and let $\,h\,$ be a nonzero number.

A difference quotient is an expression of the form $$\cssId{s9}{\frac{f(x+h)-f(x)}h}\ ,$$ which is a simplified version of: $$\cssId{s11}{\frac{f(x+h)-f(x)}{(x+h)-x}}$$

This expression gives the slope of the line through the points:

$(x\,,\,f(x))$
and
$(x+h\,,\,f(x+h))$

Since the expression is a quotient (division) of differences (subtractions), the name difference quotient is appropriate.

difference quotients, h positive

Comments on Difference Quotients

difference quotients, h negative

Example: A Difference Quotient for the Reciprocal Function

Let $\displaystyle\,f(x) = \frac 1x\,.$

Find the difference quotient $\ \frac{f(x+h)-f(x)}h\ ,$ and write it in a form with no $\,h\,$ in the denominator.

What does the difference quotient approach, as $\,h\,$ approaches zero?   What is the significance of this result?

Solution:
$\displaystyle\frac{f(x+h)-f(x)}h$ The desired difference quotient
        $\displaystyle = \frac{\frac 1{x+h} - \frac 1x}h$ Function evaluation; use definition of $\,f\,$
        $\displaystyle = \frac{\frac 1{x+h}\cdot\frac xx - \frac 1x\cdot\frac{x+h}{x+h}}h$   Get a common denominator
        $\displaystyle = \frac{\frac x{x(x+h)} - \frac {x+h}{x(x+h)}}h$ Simplify
        $\displaystyle = \frac{\frac {x - (x+h)}{x(x+h)}}h$ Write the numerator as a single fraction
        $\displaystyle = {\frac {x-x-h}{x(x+h)}}\cdot\frac 1h$ Distributive law; and, dividing by $\,h\,$ is the same as multiplying by $\,\frac 1h\,$
        $\displaystyle = {\frac {-h}{x(x+h)h}}$ $x - x = 0\,$;   and multiply across
        $\displaystyle = {\frac {-1}{x(x+h)}}$ Cancel:  since $\,h\ne 0\,,$ $\,\frac hh = 1\,.$

Now, the factor of $\,h\,$ is gone in the denominator!

As $\,h\,$ approaches zero, $\, {\frac {-1}{x(x+h)}}\,$ approaches $\,\frac{-1}{x(x+0)}\,,$ which equals $\, -\frac 1{x^2}\,.$

The formula $\,-\frac{1}{x^2}\,$ gives the slopes of the tangent lines to the graph of $\,f(x) = \frac 1x\,$!

For example, the slope of the tangent line to $\,f(x) = \frac 1x\,$ at the point $\,(3,\frac 13)\,$ is $\,-\frac{1}{3^2} = -\frac 19\,.$

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