Question

score on last attempt: 0 out of 2
score in gradebook: 0 out of 2
given that $f(x)=sqrt{x + 5}$, write an expression for the average rate of change of $f$ over the following intervals:
a. from $x = 15$ to $x = 19$
b. from $x = 19$ to $x = 19.03$

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$.

Step2: Calculate for part a

First, find $f(15)$ and $f(19)$. Given $f(x)=\sqrt{x + 5}$, then $f(15)=\sqrt{15 + 5}=\sqrt{20}=2\sqrt{5}$, and $f(19)=\sqrt{19+5}=\sqrt{24}=2\sqrt{6}$.
The average rate of change from $x = 15$ to $x = 19$ is $\frac{f(19)-f(15)}{19 - 15}=\frac{2\sqrt{6}-2\sqrt{5}}{4}=\frac{\sqrt{6}-\sqrt{5}}{2}$.

Step3: Calculate for part b

Find $f(19)$ and $f(19.03)$. $f(19)=\sqrt{19 + 5}=\sqrt{24}=2\sqrt{6}$, and $f(19.03)=\sqrt{19.03+5}=\sqrt{24.03}$.
The average rate of change from $x = 19$ to $x = 19.03$ is $\frac{f(19.03)-f(19)}{19.03 - 19}=\frac{\sqrt{24.03}-2\sqrt{6}}{0.03}$.

Answer:

a. $\frac{\sqrt{6}-\sqrt{5}}{2}$
b. $\frac{\sqrt{24.03}-2\sqrt{6}}{0.03}$