Question: Let $h(x) = \frac{f(x)}{2x}$, find $h'(2)$ where $f(x)$ is defined by the piecewise linear graph with points $(0,0)$ to $(3,9)$ and then horizontal.
1. **State the problem:** We need to find the derivative $h'(2)$ where $h(x) = \frac{f(x)}{2x}$ and $f(x)$ is given by a piecewise linear graph.
2. **Understand the function $f(x)$:** From the graph, $f(x)$ is linear from $(0,0)$ to $(3,9)$, so the slope is $\frac{9-0}{3-0} = 3$. Thus, for $0 \leq x \leq 3$, $f(x) = 3x$. For $x > 3$, $f(x) = 9$ (constant).
3. **Find $f(2)$ and $f'(2)$:** Since $2 < 3$, $f(2) = 3 \times 2 = 6$ and $f'(2) = 3$ (the slope of the linear segment).
4. **Write $h(x)$ explicitly:**
$$
h(x) = \frac{f(x)}{2x}
$$
5. **Use the quotient rule to find $h'(x)$:**
$$
h'(x) = \frac{(f'(x))(2x) - f(x)(2)}{(2x)^2} = \frac{2x f'(x) - 2 f(x)}{4x^2} = \frac{x f'(x) - f(x)}{2 x^2}
$$
6. **Evaluate $h'(2)$:**
$$
h'(2) = \frac{2 \times 3 - 6}{2 \times 2^2} = \frac{6 - 6}{8} = \frac{0}{8} = 0
$$
**Final answer:**
$$
h'(2) = 0
$$