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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 \( x \)-intercept(s) of \( f \) is \( x= \) \( \square \) (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.) B. The function \( f \) has no \( x \)-intercepts. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 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) \( \square \) (Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The function \( f \) is never increasing.

Ask by Mcdonald Cummings. in the United States
Mar 12,2025

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**Summary of \( y=f(x) \):** - **\( y \)-Intercept:** \( (0, 325) \) - **\( x \)-Intercepts:** None - **Increasing Interval:** \( (\ln(324), \infty) \) - **Graph:** Starts at \( (0, 325) \), decreases to a minimum at \( x = \ln(324) \), then increases, with both ends approaching infinity. No \( x \)-intercepts and concave upward throughout.

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To find the \( x \)-intercept(s), set \( f(x) = 0 \): \[ e^{0.5 x} + 324 e^{-0.5 x} = 0. \] This equation has no solution since both \( e^{0.5 x} \) and \( 324 e^{-0.5 x} \) are always positive. Thus, the function has no \( x \)-intercepts. So, the correct choice is: B. The function \( f \) has no \( x \)-intercepts. For the \( y \)-intercept, we evaluate \( f(0) \): \[ f(0) = e^{0} + 324 \cdot e^{0} = 1 + 324 = 325. \] So the correct choice is: A. The \( y \)-intercept of \( f \) is \( y=325 \). To determine if the function is increasing, we find its derivative: \[ f'(x) = 0.5 e^{0.5 x} - 162 e^{-0.5 x}. \] Setting \( f'(x) = 0 \) results in the critical point \( x = 0 \). Investigating the intervals around this point shows that \( f'(x) > 0 \) for all \( x \). Thus, the function is increasing everywhere. A. The function \( f \) is increasing on the subinterval(s) \( (-\infty, \infty) \).

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