1. **State the problem:** We need to find the derivative of the product function $h(x) = f(x) \cdot g(x)$ at $x=1$, $x=2$, and $x=3$. The graph of $f(x)$ has a sharp corner at $x=2$, and $g(x)$ is a straight line. 2. **Recall the product rule:** The derivative of a product of two functions is given by $$h'(x) = f'(x)g(x) + f(x)g'(x).$$ 3. **Find $f'(x)$ and $g'(x)$ at the points:** - $f(x)$ is piecewise linear: from $(0,0)$ to $(2,2)$ and from $(2,2)$ to $(4,0)$. - For $0 < x < 2$, slope $f'(x) = \frac{2-0}{2-0} = 1$. - For $2 < x < 4$, slope $f'(x) = \frac{0-2}{4-2} = -1$. - At $x=2$, $f'(2)$ does not exist due to the sharp corner. - $g(x)$ is a straight line from approximately $(0,4)$ to $(4,0)$. - Slope $g'(x) = \frac{0-4}{4-0} = -1$ everywhere. 4. **Evaluate $f(x)$ and $g(x)$ at $x=1,2,3$:** - $f(1)$ lies on the first segment: $f(1) = 1$. - $f(2) = 2$ (peak). - $f(3)$ lies on the second segment: $f(3) = 1$. - $g(1) = 4 - 1 = 3$ (since slope is -1 and intercept approx 4). - $g(2) = 4 - 2 = 2$. - $g(3) = 4 - 3 = 1$. 5. **Calculate $h'(x)$ at each point:** **A. At $x=1$:** $$h'(1) = f'(1)g(1) + f(1)g'(1) = (1)(3) + (1)(-1) = 3 - 1 = 2.$$ **B. At $x=2$:** $f'(2)$ does not exist due to the sharp corner, so $$h'(2) = \text{dne}.$$ **C. At $x=3$:** $$h'(3) = f'(3)g(3) + f(3)g'(3) = (-1)(1) + (1)(-1) = -1 - 1 = -2.$$ **Final answers:** - $h'(1) = 2$ - $h'(2) = \text{dne}$ - $h'(3) = -2$