Question

find the average rate of change of ( g(x) = \frac{-19}{x - 2} ) over the interval (-17, -16). write your answer as an integer, fraction, or decimal rounded to the nearest tenth. simplify any fractions.

Explanation:

Step1: Recall the average rate of change formula

The average rate of change of a function \( g(x) \) over the interval \([a, b]\) is given by \(\frac{g(b) - g(a)}{b - a}\). Here, \( a=-17 \), \( b = - 16\), and \( g(x)=\frac{-19}{x - 2}\).

Step2: Calculate \( g(-16) \)

Substitute \( x=-16 \) into \( g(x) \):
\( g(-16)=\frac{-19}{-16 - 2}=\frac{-19}{-18}=\frac{19}{18} \)

Step3: Calculate \( g(-17) \)

Substitute \( x = - 17\) into \( g(x) \):
\( g(-17)=\frac{-19}{-17-2}=\frac{-19}{-19} = 1\)

Step4: Calculate the average rate of change

Using the formula \(\frac{g(b)-g(a)}{b - a}\), substitute \( a=-17 \), \( b=-16 \), \( g(-16)=\frac{19}{18}\) and \( g(-17) = 1\):
\[

$$\begin{align*} \frac{g(-16)-g(-17)}{-16-(-17)}&=\frac{\frac{19}{18}-1}{-16 + 17}\\ &=\frac{\frac{19-18}{18}}{1}\\ &=\frac{\frac{1}{18}}{1}\\ &=\frac{1}{18}\approx0.055\cdots\approx0.1 \end{align*}$$

\]

Answer:

\(\frac{1}{18}\) (or approximately \(0.1\))