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Exercises Limits and derivatives

Example 1 from Precalculus, 12.4 Derivatives

Finding the Average Rate of Change

Find the average rate of change connecting the points $(2, −6)$ and $(−1, 5) .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
We know the average rate of change connecting two points may be given by

$AROC = \frac{f (a + h) - f (a)}{h} .$

If one point is $(2, - 6),$ or $(2, f (2)),$ then $f (2) = −6.$
The value $h$ is the displacement from $2$ to $- 1,$ which equals $- 1 - 2 = −3.$
For the other point, $f (a + h)$ is the y-coordinate at $a + h,$ which is $2 + (−3)$ or $−1,$ so $f (a + h) = f (−1) = 5.$
$\begin{array}{l} AROC = \frac{f (a + h) - f (a)}{h} \\ = \frac{5 - (- 6)}{- 3} \\ = \frac{11}{- 3} \\ = - \frac{11}{3} \end{array}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Find the average rate of change connecting the points $(- 5, 1.5)$ and $(- 2.5, 9) .$

How did it go?

PreviousDetermining Whether a Piecewise Function Is Continuous Next Finding the Derivative of a Polynomial Function

From Precalculus (1st edition) by OpenStax (Rice University), section 12.4, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.