1. **State the problem:** We need to graph the function $f$ over the interval $[-5,6]$. Given: - $f$ is piecewise linear (made of closed line segments joined end to end). - $f(-5) = 1$. - The derivative $f'$ is a step function defined as: - $f'(x) = 0$ for $x \in [-5,-5)$ (not relevant since starting at -5), and $x \in [-7,-5)$ - $f'(x) = 1$ for $x \in [-5,-3)$ - $f'(x) = 0$ for $x \in [-3,-1)$ - $f'(x) = -1$ for $x \in [-1,2)$ - $f'(x) = 0$ for $x \in [2,7)$ 2. **Formula and rules:** Since $f'$ is the derivative of $f$, the slope of $f$ on each interval equals the value of $f'$ on that interval. 3. **Calculate $f$ on each interval using the slope and initial point:** - On $[-5,-3]$, slope $m=1$: $$f(x) = f(-5) + 1 \cdot (x + 5) = 1 + (x + 5) = x + 6$$ At $x=-3$, $$f(-3) = -3 + 6 = 3$$ - On $[-3,-1]$, slope $m=0$ (constant): $$f(x) = f(-3) = 3$$ At $x=-1$, $$f(-1) = 3$$ - On $[-1,2]$, slope $m=-1$: $$f(x) = f(-1) - 1 \cdot (x + 1) = 3 - (x + 1) = 2 - x$$ At $x=2$, $$f(2) = 2 - 2 = 0$$ - On $[2,6]$, slope $m=0$ (constant): $$f(x) = f(2) = 0$$ 4. **Summary of points and segments:** - $(-5,1)$ to $(-3,3)$ with slope 1 - $(-3,3)$ to $(-1,3)$ with slope 0 - $(-1,3)$ to $(2,0)$ with slope -1 - $(2,0)$ to $(6,0)$ with slope 0 5. **Final piecewise function:** $$ f(x) = \begin{cases} x + 6 & -5 \leq x \leq -3 \\ 3 & -3 < x \leq -1 \\ 2 - x & -1 < x \leq 2 \\ 0 & 2 < x \leq 6 \end{cases} $$ This matches the graph with vertices at $(-5,1)$, $(-3,3)$, $(-1,3)$, $(2,0)$, and $(6,0)$ connected by line segments. **Answer:** The graph of $f$ is piecewise linear with the above points and slopes.