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: Since the expression is a quotient (division) of differences (subtractions), the name difference quotient is appropriate. |
![]() |
|
![]() |
$\displaystyle\frac{f(x+h)-f(x)}h$ | 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,
$\,\displaystyle {\frac {-1}{x(x+h)}}\,$ approaches $\,\displaystyle\frac{-1}{x(x+0)}\,$,
which equals $\,\displaystyle -\frac 1{x^2}\,$. The formula $\displaystyle\,-\frac{1}{x^2}\,$ gives the slopes of the tangent lines to the graph of $\displaystyle\,f(x) = \frac 1x\,$! For example, the slope of the tangent line to $\displaystyle\,f(x) = \frac 1x\,$ at the point $\displaystyle\,(3,\frac 13)\,$ is $\displaystyle\,-\frac{1}{3^2} = -\frac 19\,$. Click ‘Submit’ on the WolframAlpha widget below to explore the power of Wolfram Alpha! You can create your own widgets by clicking here! |
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
PROBLEM TYPES:
|