Chapter 5.7, Problem 7E

Solution

To determine: The average rate of change over the given intervals for the function $f\left(x\right)={\log }_{1/2}\left(x-3\right)$ and its effect on graph of the function if $x$ increases.

The graph falls more and more steeply as $x$ increases.

Given information:

The function is $f\left(x\right)={\log }_{1/2}\left(x-3\right)$ and the intervals are $\left[4,5\right],\left[5,7\right],\left[7,11\right],\text{and }\left[11,19\right]$.

Formula used:

The average rate of change over an interval $\left[x_1,x_2\right]$ is $\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}$.

Explanation:

Calculate the average rate of change in the given function over the given intervals.

Interval	Average Rate of Change
$\left[4,5\right]$	$\frac{\left[{\log }_{1/2}\left(5-3\right)\right]-\left[{\log }_{1/2}\left(4-3\right)\right]}{5-4}=-1$
$\left[5,7\right]$	$\frac{\left[{\log }_{1/2}\left(7-3\right)\right]-\left[{\log }_{1/2}\left(5-3\right)\right]}{7-5}=-\frac{1}{2}$
$\left[7,11\right]$	$\frac{\left[{\log }_{1/2}\left(11-3\right)\right]-\left[{\log }_{1/2}\left(7-3\right)\right]}{11-7}=-\frac{1}{4}$
$\left[11,19\right]$	$\frac{\left[{\log }_{1/2}\left(19-3\right)\right]-\left[{\log }_{1/2}\left(11-3\right)\right]}{19-11}=-\frac{1}{8}$

The average rate of change is always negative, so the graph will always fall. The average rate of change decreases over each interval is equivalent to the slope of the segment from $\left(x_1,f\left(x_1\right)\right)$ to $\left(x_2,f\left(x_2\right)\right)$ , the graph falls more and more steeply as $x$ increases.