Question

1. find aroc 0,3

$f(x)$
$g(x) = x^2 - 2x$
$f(0)=0$
$f(3)=2$
$\frac{2 - 0}{3 - 0}=\frac{2}{3}$
$g(0)=0$
$f(x)$ has a greater aroc than $g(x)$ $square$

Explanation:

Step1: Recall AROC formula

The average rate of change (AROC) of a function $h(x)$ on $[a,b]$ is $\frac{h(b)-h(a)}{b-a}$.

Step2: Calculate AROC for $f(x)$

From the graph: $f(0)=1$, $f(3)=2$.
$\text{AROC of } f(x) = \frac{f(3)-f(0)}{3-0} = \frac{2-1}{3-0} = \frac{1}{3}$

Step3: Calculate AROC for $g(x)$

Given $g(x)=x^2-2x$:
$g(0)=0^2-2(0)=0$, $g(3)=3^2-2(3)=9-6=3$.
$\text{AROC of } g(x) = \frac{g(3)-g(0)}{3-0} = \frac{3-0}{3-0} = 1$

Step4: Compare the two AROCs

$1 > \frac{1}{3}$, so $g(x)$ has a greater AROC.

Answer:

The AROC of $f(x)$ on $[0,3]$ is $\frac{1}{3}$, the AROC of $g(x)$ on $[0,3]$ is $1$. $g(x)$ has a greater AROC than $f(x)$, so the box next to "$g(x)$ has a greater AROC than $f(x)$" should be checked.