1. **Problem statement:** Consider the function $h(x) = e^{-g(x)}$ where $g(x)$ is continuous with a continuous first derivative. If $g(x)$ has a local maximum at the point $a$, what can we say about $h(x)$ at $a$? 2. **Recall the definitions and formulas:** - A local maximum of $g(x)$ at $a$ means $g'(a) = 0$ and $g''(a) < 0$. - The function $h(x) = e^{-g(x)}$ is a composition of exponential and $-g(x)$. 3. **Find the first derivative of $h(x)$:** $$h'(x) = \frac{d}{dx} e^{-g(x)} = e^{-g(x)} \cdot (-g'(x)) = -g'(x) e^{-g(x)}$$ 4. **Evaluate $h'(a)$:** Since $g'(a) = 0$ (because $g$ has a local max at $a$), $$h'(a) = -0 \cdot e^{-g(a)} = 0$$ 5. **Find the second derivative of $h(x)$:** Using the product rule, $$h''(x) = \frac{d}{dx}(-g'(x) e^{-g(x)}) = -g''(x) e^{-g(x)} + g'(x)^2 e^{-g(x)}$$ 6. **Evaluate $h''(a)$:** Since $g'(a) = 0$, $$h''(a) = -g''(a) e^{-g(a)} + 0 = -g''(a) e^{-g(a)}$$ 7. **Analyze the sign of $h''(a)$:** - $g''(a) < 0$ because $g$ has a local maximum at $a$. - $e^{-g(a)} > 0$ for all real $g(a)$. Therefore, $$h''(a) = -g''(a) e^{-g(a)} > 0$$ 8. **Conclusion:** Since $h'(a) = 0$ and $h''(a) > 0$, $h(x)$ has a local minimum at $a$. **Answer:** A. $h(x)$ has a local minimum at the point $a$. --- **Problem 2 (not solved here, only counted):** If $f$ is continuous on $[1,5]$ with $f(1) = 5$ and $f(5) = 2$, is there a $c$ in $(1,5)$ with $f(c) = 1$ by the Intermediate Value Theorem? **q_count:** 2