Question

1. let $f(x)=sqrt{x}$. find a number $b$ so that the average rate of change of $f$ on the interval $1,b$ is $\frac{1}{7}$.

Explanation:

Step1: Recall average rate - of - change formula

The average rate of change of $y = f(x)$ on $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 1$, $f(x)=\sqrt{x}$, so $\frac{f(b)-f(1)}{b - 1}=\frac{1}{7}$.

Step2: Substitute function values

Since $f(1)=\sqrt{1}=1$ and $f(b)=\sqrt{b}$, we have $\frac{\sqrt{b}-1}{b - 1}=\frac{1}{7}$. Factor the denominator $b - 1=(\sqrt{b}-1)(\sqrt{b}+1)$. Then $\frac{\sqrt{b}-1}{(\sqrt{b}-1)(\sqrt{b}+1)}=\frac{1}{7}$.

Step3: Simplify and solve

Cancel out $\sqrt{b}-1$ (assuming $\sqrt{b}
eq1$ i.e., $b
eq1$), we get $\frac{1}{\sqrt{b}+1}=\frac{1}{7}$. Cross - multiply to obtain $\sqrt{b}+1 = 7$, so $\sqrt{b}=6$ and $b = 36$.

Answer:

$36$