1. **Problem statement:** Given a piecewise linear function $f$ defined by points $(-4,-3)$, $(-1,-2)$, $(1,2)$, $(2,0)$, and $(3,-2)$, define $g(x) = \int_{-4}^x f(t) \, dt$. Find $g(-1)$, $g'(-1)$, and $g''(-1)$. 2. **Recall the Fundamental Theorem of Calculus:** - $g'(x) = f(x)$ because $g$ is the integral of $f$. - $g''(x) = f'(x)$, the derivative of $f$. 3. **Calculate $g(-1)$:** $g(-1) = \int_{-4}^{-1} f(t) \, dt$. Since $f$ is linear between $-4$ and $-1$, find the equation of $f$ on $[-4,-1]$: Slope $m = \frac{-2 - (-3)}{-1 - (-4)} = \frac{1}{3}$. Equation: $f(t) = \frac{1}{3}(t + 4) - 3 = \frac{1}{3}t + \frac{4}{3} - 3 = \frac{1}{3}t - \frac{5}{3}$. Integrate: $$g(-1) = \int_{-4}^{-1} \left(\frac{1}{3}t - \frac{5}{3}\right) dt = \left[ \frac{1}{6}t^2 - \frac{5}{3}t \right]_{-4}^{-1}$$ Calculate: At $t=-1$: $\frac{1}{6}(-1)^2 - \frac{5}{3}(-1) = \frac{1}{6} + \frac{5}{3} = \frac{1}{6} + \frac{10}{6} = \frac{11}{6}$. At $t=-4$: $\frac{1}{6}(-4)^2 - \frac{5}{3}(-4) = \frac{16}{6} + \frac{20}{3} = \frac{16}{6} + \frac{40}{6} = \frac{56}{6} = \frac{28}{3}$. So, $$g(-1) = \frac{11}{6} - \frac{28}{3} = \frac{11}{6} - \frac{56}{6} = -\frac{45}{6} = -\frac{15}{2} = -7.5$$ 4. **Calculate $g'(-1)$:** By the Fundamental Theorem of Calculus, $g'(-1) = f(-1)$. From the graph, $f(-1) = -2$. 5. **Calculate $g''(-1)$:** $g''(x) = f'(x)$. Since $f$ is piecewise linear, $f'$ is constant on intervals but may be undefined at points where $f$ changes slope. Check slopes around $x=-1$: - On $(-4,-1)$, slope $m_1 = \frac{1}{3}$. - On $(-1,1)$, slope $m_2 = \frac{2 - (-2)}{1 - (-1)} = \frac{4}{2} = 2$. Since slopes differ, $f'$ is not defined at $x=-1$. **Final answers:** $$g(-1) = -\frac{15}{2}$$ $$g'(-1) = -2$$ $$g''(-1) \text{ does not exist}$$