1. **Problem Statement:** Given $g(x) = \int_0^x f(t) \, dt$ where $f$ is a piecewise linear function with points $(0,0), (1,1), (3,4), (6,1), (9,-1), (10,1)$. (a) Use the Fundamental Theorem of Calculus Part 1 to graph $g'$. (b) Find $g(3)$, $g'(3)$, and $g''(3)$. (c) Determine if $g$ has a local max, min, or neither at $x=6$. (d) Determine if $g$ has a local max, min, or neither at $x=9$. 2. **Fundamental Theorem of Calculus Part 1:** It states that if $g(x) = \int_0^x f(t) \, dt$, then $g'(x) = f(x)$. This means the derivative of $g$ is the original function $f$. 3. **Step (a): Graph of $g'$** Since $g'(x) = f(x)$, the graph of $g'$ is exactly the graph of $f$ given. 4. **Step (b): Calculate $g(3)$, $g'(3)$, and $g''(3)$** - $g(3) = \int_0^3 f(t) \, dt$ is the area under $f$ from 0 to 3. - $g'(3) = f(3)$ by the Fundamental Theorem. - $g''(3) = f'(3)$ since $g''(x) = f'(x)$. Calculate $g(3)$: - From 0 to 1, $f(t)$ rises linearly from 0 to 1, area is a triangle with base 1 and height 1: $\frac{1}{2} \times 1 \times 1 = 0.5$. - From 1 to 3, $f(t)$ rises linearly from 1 to 4, area is a trapezoid with bases 1 and 4 and height 2: $\frac{1+4}{2} \times 2 = 5$. Total area $g(3) = 0.5 + 5 = 5.5$. Calculate $g'(3)$: - $g'(3) = f(3) = 4$ (given point). Calculate $g''(3)$: - $f$ is piecewise linear, slope from 1 to 3 is $\frac{4-1}{3-1} = \frac{3}{2} = 1.5$. - So $g''(3) = f'(3) = 1.5$. 5. **Step (c): Local max/min at $x=6$ for $g$** - $g'(6) = f(6) = 1$ (from graph). - $g''(6) = f'(6)$ is slope of $f$ at 6. - Between 3 and 6, slope of $f$ is $\frac{1-4}{6-3} = \frac{-3}{3} = -1$. - Between 6 and 9, slope of $f$ is $\frac{-1-1}{9-6} = \frac{-2}{3} \approx -0.67$. - Since $f$ is decreasing around 6, $f'(6)$ is negative. - $g'(6) = 1 > 0$ and $g''(6) < 0$ means $g$ is increasing but concave down, so no local max or min at 6. 6. **Step (d): Local max/min at $x=9$ for $g$** - $g'(9) = f(9) = -1$. - Slope of $f$ from 6 to 9 is $-0.67$ (negative), from 9 to 10 is $\frac{1 - (-1)}{10 - 9} = 2$ (positive). - So $f'(9)$ changes from negative to positive, meaning $g''(9)$ changes from negative to positive. - $g'(9) = -1 < 0$ and $g''(9)$ changes sign from negative to positive, indicating $g$ has a local minimum at $x=9$. **Final answers:** - (a) $g'(x) = f(x)$, graph of $g'$ is the given graph of $f$. - (b) $g(3) = 5.5$, $g'(3) = 4$, $g''(3) = 1.5$. - (c) At $x=6$, $g$ has neither a local max nor a local min. - (d) At $x=9$, $g$ has a local minimum.