Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of \( y=f(x) \). \( f(x)=e^{0.5 x}+324 e^{-0.5 x} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice A. The domain of \( f \) is (Type your answer in interval notation. Use a comma to separate answers as needed.) B. The domain of \( f \) is empty.

1. We begin with the function \[ f(x)=e^{0.5x}+324e^{-0.5x}. \] 2. **Domain:** Since both exponentials \(e^{0.5x}\) and \(e^{-0.5x}\) are defined for all real numbers, the domain of \(f\) is \[ (-\infty,\infty). \] 3. **End Behavior:** - As \( x\to\infty \), the term \(e^{0.5x}\) becomes dominant and tends to infinity, so \[ \lim_{x\to\infty} f(x) = \infty. \] - As \( x\to -\infty \), the term \(e^{-0.5x}\) becomes dominant and tends to infinity, thus \[ \lim_{x\to -\infty} f(x) = \infty. \] 4. **Rewriting the Function (for minimum value determination):** Let \[ u=e^{0.5x}, \] so that \( e^{-0.5x}=\frac{1}{u} \) and the function becomes \[ f(u)= u + \frac{324}{u}. \] 5. **Finding the Minimum Value:** To minimize \[ g(u) = u + \frac{324}{u}, \] for \(u>0\), we can use the AM–GM inequality: \[ \frac{u + \dfrac{324}{u}}{2} \ge \sqrt{u\cdot\frac{324}{u}} = \sqrt{324}=18. \] Equality holds when \[ u=\frac{324}{u} \quad\Longrightarrow\quad u^2=324 \quad\Longrightarrow\quad u=18. \] Since \(u=e^{0.5x}\), setting \(e^{0.5x}=18\) gives \[ 0.5x=\ln(18)\quad\Longrightarrow\quad x=2\ln(18). \] Therefore, the minimum value of \(f\) is \[ f(x)=18+\frac{324}{18}=18+18=36, \] and it occurs at \( x=2\ln(18) \). 6. **Sketch Summary:** - The domain of \(f\) is \((-\infty,\infty)\). - \(f(x)\) tends to infinity as \(x\to\infty\) and as \(x\to -\infty\). - The graph has a single minimum point at \(\left(2\ln(18),\,36\right)\). - The \(y\)-intercept is obtained by evaluating \(f(0)=e^0+324e^0=1+324=325\). **Answer:** A. The domain of \(f\) is \((-\infty,\infty)\).