Question

1. find the average rate of change of the function over the given interval.

a. $y = \ln(5 - x)$; over $2,4$
b. $y = e^{x - 3}$; over $3,7$
c. $y = \sqrt{6 - 3x}$; over $0,2$

1. find the limit.

Explanation:

a.

Step1: Recall average rate of change formula

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here $f(x)=\ln(5 - x)$, $a = 2$, $b=4$.

Step2: Calculate $f(4)$ and $f(2)$

$f(4)=\ln(5 - 4)=\ln(1)=0$, $f(2)=\ln(5 - 2)=\ln(3)$.

Step3: Compute average rate of change

$\frac{f(4)-f(2)}{4 - 2}=\frac{0-\ln(3)}{2}=-\frac{\ln(3)}{2}$.

b.

Step1: Recall average rate of change formula

For $y = f(x)=e^{x - 3}$ over $[a,b]=[3,7]$, the average rate of change is $\frac{f(b)-f(a)}{b - a}$.

Step2: Calculate $f(7)$ and $f(3)$

$f(7)=e^{7 - 3}=e^{4}$, $f(3)=e^{3 - 3}=e^{0}=1$.

Step3: Compute average rate of change

$\frac{f(7)-f(3)}{7 - 3}=\frac{e^{4}-1}{4}$.

c.

Step1: Recall average rate of change formula

For $y = f(x)=\sqrt{6-3x}$ over $[a,b]=[0,2]$, the average rate of change is $\frac{f(b)-f(a)}{b - a}$.

Step2: Calculate $f(2)$ and $f(0)$

$f(2)=\sqrt{6-3\times2}=\sqrt{6 - 6}=0$, $f(0)=\sqrt{6-3\times0}=\sqrt{6}$.

Step3: Compute average rate of change

$\frac{f(2)-f(0)}{2 - 0}=\frac{0-\sqrt{6}}{2}=-\frac{\sqrt{6}}{2}$.

Answer:

a. $-\frac{\ln(3)}{2}$
b. $\frac{e^{4}-1}{4}$
c. $-\frac{\sqrt{6}}{2}$