Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of \( \mathrm{y}=\mathrm{f}(\mathrm{x}) \). \( \mathrm{f}(\mathrm{x})=e^{0.5 \mathrm{x}}+324 e^{-0.5 \mathrm{x}} \) A. The function f has a local maximum at \( \mathrm{x}= \) (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.). B. The function f has no local maximum. A. The function f has a local minimum at \( \mathrm{x}= \) In (324). (Type an exact answer. Use a comma to separate answers as needed.) B. The function f has no local minimum. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) is concave upward on the subinterval(s) (Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The function \( f \) is never concave upward.

**Step 1. Find the Critical Points** We start with the function \[ f(x)=e^{0.5x}+324e^{-0.5x}. \] Differentiate with respect to \(x\): \[ f'(x)=0.5e^{0.5x}-324(0.5)e^{-0.5x}=\frac{1}{2}\Bigl(e^{0.5x}-324\,e^{-0.5x}\Bigr). \] Set \(f'(x)=0\) to find the critical points: \[ \frac{1}{2}\Bigl(e^{0.5x}-324\,e^{-0.5x}\Bigr) = 0 \quad\Longrightarrow\quad e^{0.5x} -324\,e^{-0.5x} = 0. \] Rearrange: \[ e^{0.5x} = 324\,e^{-0.5x}. \] Multiply both sides by \(e^{0.5x}\): \[ e^{0.5x}\cdot e^{0.5x} = 324 \quad\Longrightarrow\quad e^{x}=324. \] Taking natural logarithms on both sides gives: \[ x=\ln(324). \] **Step 2. Determine the Type of Critical Point** To decide whether \(x=\ln(324)\) is a maximum or a minimum, we compute the second derivative. Differentiate \(f'(x)\): \[ f'(x)=\frac{1}{2}\Bigl(e^{0.5x}-324\,e^{-0.5x}\Bigr). \] Differentiate term‐by‐term: \[ f''(x)=\frac{1}{2}\Bigl(0.5e^{0.5x}+324\cdot 0.5\,e^{-0.5x}\Bigr) =\frac{1}{4}\Bigl(e^{0.5x}+324\,e^{-0.5x}\Bigr). \] Since \(e^{0.5x}>0\) and \(324\,e^{-0.5x}>0\) for all \(x\), we have: \[ f''(x)>0 \quad \text{for all } x. \] A positive second derivative indicates that the function is concave upward and thus the critical point at \(x=\ln(324)\) is a local minimum. **Step 3. Analyze Concavity** As derived, \[ f''(x)=\frac{1}{4}\Bigl(e^{0.5x}+324\,e^{-0.5x}\Bigr)>0\quad \text{for all } x. \] Therefore, the function is concave upward on the entire real line: \[ (-\infty,\,\infty). \] **Final Answers** - **Local Maximum:** B. The function \(f\) has no local maximum. - **Local Minimum:** A. The function \(f\) has a local minimum at \(x=\ln(324)\). - **Concavity:** A. The function \(f\) is concave upward on the subinterval(s) \((-\infty,\,\infty)\).