Question

find the average rate of change of the function $h(x)=\frac{3}{x + 7}$ on the interval $1,8$.

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \(y = h(x)\) on the interval \([a,b]\) is \(\frac{h(b)-h(a)}{b - a}\). Here \(a = 1\), \(b=8\), and \(h(x)=\frac{3}{x + 7}\).

Step2: Calculate \(h(8)\) and \(h(1)\)

• For \(h(8)\):

Substitute \(x = 8\) into \(h(x)\), \(h(8)=\frac{3}{8+7}=\frac{3}{15}=\frac{1}{5}\).

• For \(h(1)\):

Substitute \(x = 1\) into \(h(x)\), \(h(1)=\frac{3}{1 + 7}=\frac{3}{8}\).

Step3: Substitute into the average - rate - of - change formula

\(\frac{h(8)-h(1)}{8 - 1}=\frac{\frac{1}{5}-\frac{3}{8}}{7}\).
First, find a common denominator for the numerator: \(\frac{1\times8}{5\times8}-\frac{3\times5}{8\times5}=\frac{8}{40}-\frac{15}{40}=\frac{8 - 15}{40}=-\frac{7}{40}\).
Then, \(\frac{-\frac{7}{40}}{7}=-\frac{7}{40}\times\frac{1}{7}\).

Answer:

\(-\frac{1}{40}\)