Question

PLEASE HELP 99 POINTS

Answer 1

Presumably, $h(x)=3(5)^x$. In that case,

(A)
The average rate of change over the interval $0\le x\le1$ is

$\dfrac{h(1)-h(0)}{1-0}=\dfrac{15-3}1=12$

and over $2\le x\le3$, it's

$\dfrac{h(3)-h(2)}{3-2}=\dfrac{375-75}1=300$

(B)
$\dfrac{300}{12}=25$, i.e. the average rate of change over the second interval is 25 times higher. That's to be expected; $3(5)^x$ is an exponential function. As $x$ gets larger, the rate of change of $h(x)$ gets larger too.