Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of \( y=f(x) \). \( f(x)=e^{0.5 x}+324 e^{-0.5 x} \) A. The y-intercept of \( f \) is \( y=325 \) (Type an exact answer, using radicals as needed.) B. The function \( f \) has no \( y \)-intercept Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) is increasing on the subinterval(s) (In (324), 0 ). B. The function \( f \) is never increasing. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) is decreasing 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 decreasing.

\[ \textbf{1. Domain:} \quad (-\infty, \infty) \] \[ \textbf{2. Y-intercept:} \] Evaluate \[ f(0)=e^{0.5\cdot0}+324\,e^{-0.5\cdot0}=1+324=325. \] Thus, the y-intercept is \((0,325)\). \[ \textbf{3. Finding critical points:} \] Differentiate \[ f'(x)=\frac{d}{dx}\left(e^{0.5x}+324\,e^{-0.5x}\right)=0.5\,e^{0.5x}-0.5\cdot324\,e^{-0.5x}. \] Factor out \(0.5\): \[ f'(x)=0.5\left(e^{0.5x}-324\,e^{-0.5x}\right). \] Set \(f'(x)=0\): \[ e^{0.5x}-324\,e^{-0.5x}=0 \quad \Longrightarrow \quad e^{0.5x}=324\,e^{-0.5x}. \] Multiply both sides by \(e^{0.5x}\): \[ e^{0.5x}\cdot e^{0.5x}=324 \quad \Longrightarrow \quad e^{x}=324. \] Taking the natural logarithm: \[ x=\ln(324). \] \[ \textbf{4. Monotonicity:} \] For \(x<\ln(324)\): - Choose a test value \(x<\ln(324)\). Then \(e^{0.5x} < 324\,e^{-0.5x}\). Hence, \(f'(x)<0\) and \(f\) is decreasing on \[ (-\infty,\ln(324)). \] For \(x>\ln(324)\): - \(e^{0.5x} > 324\,e^{-0.5x}\), so \(f'(x)>0\) and \(f\) is increasing on \[ (\ln(324),\infty). \] \[ \textbf{5. Minimum value:} \] Since \(f'\) changes from negative to positive at \(x=\ln(324)\), there is a local (and global) minimum at this point. Evaluate: \[ f(\ln(324))=e^{0.5\ln(324)}+324\,e^{-0.5\ln(324)}. \] Recall that \(e^{0.5\ln(324)}=324^{0.5}=\sqrt{324}=18\) and \[ e^{-0.5\ln(324)}=\frac{1}{18}. \] Thus, \[ f(\ln(324))=18+324\left(\frac{1}{18}\right)=18+18=36. \] So the minimum point is \[ \left(\ln(324),\,36\right). \] \[ \textbf{6. End Behavior:} \] - As \(x\to\infty\), the term \(e^{0.5x}\) dominates, so \(f(x)\to\infty\). - As \(x\to-\infty\), the term \(324\,e^{-0.5x}\) becomes large, thus \(f(x)\to\infty\). \[ \textbf{Summary of Answers:} \] \[ \textbf{A. Y-intercept:} \quad (0,325). \] \[ \textbf{B. Increasing on:} \quad (\ln(324),\infty). \] \[ \textbf{C. Decreasing on:} \quad (-\infty,\ln(324)). \] \[ \textbf{Graph Sketch Description:} \] - The graph has a \(y\)-intercept at \((0,325)\). - It decreases on the interval \((-\infty,\ln(324))\) and reaches a minimum at \(\bigl(\ln(324),36\bigr)\). - Then, it increases on the interval \((\ln(324),\infty)\). - The end behavior is such that \(f(x)\to\infty\) as \(x\to\infty\) and \(f(x)\to\infty\) as \(x\to-\infty\). This completes the analysis and the sketching strategy for the function \( f(x)=e^{0.5 x}+324\,e^{-0.5 x} \).