Answer
The function $f$ has a greater average rate of change between $x=0$ and $x=1$. The function $g$ has a greater average rate of change between $x=1$ and $x=2$. The functions $f$ and $g$ has the same average rate of change between $x=0$ and $x=1.5$.
Work Step by Step
The average rate of change between $x=0$ and another $x$ value for $f(x)$ and $g(x)$ are:
average rate of change for $f(x)$ $= \frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}$
average rate of change for $g(x)$ $= \frac{g(x)-g(0)}{x-0}=\frac{g(x)}{x}$
To fill in the first blank, we need to plug $x=1$ into the above equations. The average rate of change for $f(x)$ and $g(x)$ therefore becomes $f(1)$ and $g(1)$, respectively. Since $f(1) > g(1)$ on the graph, the function $f$ has the larger average rate of change between $x=0$ and $x=1$. The first blank is $f$.
To fill in the second blank, we need to plug $x=2$ into the above equations. The average rate of change for $f(x)$ and $g(x)$ therefore becomes $f(2)$ and $g(2)$, respectively. Since $f(2) < g(2)$ on the graph, the function $g$ has the larger average rate of change between $x=0$ and $x=2$. The second blank is $g$.
The value of $x$ that is needed in the third blank must satisfy the following equation:
$\frac{f(x)}{x}=\frac{g(x)}{x}$
$f(x)=g(x)$
$f(x)$ is only equal to $g(x)$ when the graphs intercept each other. Since the graphs intersect at $x=1.5$, the third blank is $1.5$.
