1. **State the problem:** We are given the derivative $f'(x)$ as a step function over the interval $[-2,5]$ and the initial value $f(-2)=3$. We need to recover the function $f(x)$ and graph it over $[-2,5]$. 2. **Analyze the derivative $f'(x)$:** - For $x \in [-2,0)$, $f'(x) = -2$. - For $x \in [0,1)$, $f'(x) = 0$. - For $x \in [1,3)$, $f'(x) = 1$. - For $x \in [3,5]$, $f'(x) = -1$. 3. **Recover $f(x)$ by integrating $f'(x)$ piecewise:** Since $f'(x)$ is constant on each interval, $f(x)$ is linear on each interval with slope equal to $f'(x)$. 4. **Calculate $f(x)$ on each interval using the initial condition $f(-2)=3$: ** - On $[-2,0]$: slope $m = -2$, so $$f(x) = f(-2) + \int_{-2}^x -2 \, dt = 3 - 2(x + 2) = -2x -1.$$ Check at $x=0$: $f(0) = -2(0) -1 = -1$. - On $[0,1]$: slope $m=0$, so $f(x)$ is constant: $$f(x) = f(0) = -1.$$ - On $[1,3]$: slope $m=1$, so $$f(x) = f(1) + \int_1^x 1 \, dt = -1 + (x - 1) = x - 2.$$ Check at $x=3$: $f(3) = 3 - 2 = 1$. - On $[3,5]$: slope $m = -1$, so $$f(x) = f(3) + \int_3^x -1 \, dt = 1 - (x - 3) = 4 - x.$$ Check at $x=5$: $f(5) = 4 - 5 = -1$. 5. **Summary of $f(x)$:** $$ f(x) = \begin{cases} -2x - 1 & -2 \leq x < 0 \\ -1 & 0 \leq x < 1 \\ x - 2 & 1 \leq x < 3 \\ 4 - x & 3 \leq x \leq 5 \end{cases} $$ 6. **Interpretation:** The function $f$ is piecewise linear with slopes given by $f'(x)$ and passes through the given initial point. The graph consists of connected line segments with vertices at $(-2,3)$, $(0,-1)$, $(1,-1)$, $(3,1)$, and $(5,-1)$. **Final answer:** The recovered function $f(x)$ is the piecewise linear function above.