1. Consider the function $h(x) = e^{-g(x)}$ where $g(x)$ is continuous with a continuous first derivative and has a local maximum at $x=a$. 2. Since $g(x)$ has a local maximum at $a$, we know that $g'(a) = 0$ and $g''(a) < 0$. 3. To analyze $h(x)$ at $a$, compute its first derivative: $$h'(x) = \frac{d}{dx} e^{-g(x)} = e^{-g(x)} \cdot (-g'(x)) = -g'(x) e^{-g(x)}$$ 4. At $x=a$, since $g'(a) = 0$, we have: $$h'(a) = -g'(a) e^{-g(a)} = 0$$ 5. Next, compute the second derivative of $h(x)$: $$h''(x) = \frac{d}{dx} h'(x) = \frac{d}{dx} \left(-g'(x) e^{-g(x)}\right) = -g''(x) e^{-g(x)} + g'(x)^2 e^{-g(x)}$$ 6. At $x=a$, since $g'(a) = 0$, this simplifies to: $$h''(a) = -g''(a) e^{-g(a)}$$ 7. Because $g''(a) < 0$ and $e^{-g(a)} > 0$, it follows that: $$h''(a) = -g''(a) e^{-g(a)} > 0$$ 8. A positive second derivative at $a$ means $h(x)$ has a local minimum at $a$. **Answer for problem 2:** A. has a local minimum at the point $a$. --- 1. Given $f$ is continuous on $[1,5]$ with $f(1) = 5$ and $f(5) = 2$, and the Intermediate Value Theorem (IVT) states that for any value between $f(1)$ and $f(5)$, there exists some $c$ in $(1,5)$ such that $f(c)$ equals that value. 2. Since $1$ is not between $5$ and $2$ (because $1 < 2 < 5$ is false), the IVT does not guarantee a $c$ with $f(c) = 1$. 3. Therefore, the statement "there is a $c$ in $(1,5)$ with $f(c) = 1$" is false. **Answer for problem 3:** B. False