Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of \( y=f(x) \). \( f(x)=e^{0.5 x}+256 e^{-0.5 x} \) 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 \( x= \) (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The \( y \)-intercept of \( f \) is \( y=257 \). The function \( f \) has no \( y \)-intercept. (Type an exact answer, using radicals as needed.) 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) \( \square \). B. The . The function \( f \) is never increasing. B.

1. Write the function in its given form: \[ f(x)=e^{0.5x}+256e^{-0.5x} \] 2. **Find the \(x\)-intercepts:** To find the \(x\)-intercepts we set \( f(x)=0 \): \[ e^{0.5x}+256e^{-0.5x}=0. \] Since both \( e^{0.5x} \) and \( 256e^{-0.5x} \) are positive for all real \( x \), their sum is always positive. Therefore, there are no solutions and hence no \(x\)-intercepts. 3. **Find the \(y\)-intercept:** Evaluate \( f(x) \) at \( x=0 \): \[ f(0)=e^0+256e^0=1+256=257. \] So the \(y\)-intercept is \( y=257 \). 4. **Find where \( f \) is increasing or decreasing:** First, compute the first derivative: \[ f'(x)=\frac{d}{dx}\left(e^{0.5x}\right)+\frac{d}{dx}\left(256e^{-0.5x}\right) =0.5e^{0.5x}-128e^{-0.5x}. \] To find the critical point, set \( f'(x)=0 \): \[ 0.5e^{0.5x}-128e^{-0.5x}=0. \] Rearranging, \[ 0.5e^{0.5x}=128e^{-0.5x}. \] Multiply both sides by \( e^{0.5x} \) to obtain: \[ 0.5e^{x}=128. \] Solve for \( e^x \): \[ e^x=256. \] Taking natural logarithms gives: \[ x=\ln(256). \] Recognizing that \( 256=2^8 \), we can write: \[ x=8\ln2. \] Next, determine the sign of \( f'(x) \): - For \( x < 8\ln2 \): Choosing a test value (for example, \( x=0 \)): \[ f'(0)=0.5e^0-128e^0=0.5-128<0, \] so \( f \) is decreasing on \( (-\infty,\,8\ln2) \). - For \( x > 8\ln2 \): \( f'(x)>0 \) so \( f \) is increasing on \( (8\ln2,\,\infty) \). 5. **Determine the minimum value:** Since the function decreases then increases, it has a global minimum at \( x=8\ln2 \). Compute \( f(8\ln2) \): \[ f(8\ln2)=e^{0.5(8\ln2)}+256e^{-0.5(8\ln2)} =e^{4\ln2}+256e^{-4\ln2}. \] Recognize that \( e^{4\ln2}=2^4=16 \) and \( e^{-4\ln2}=2^{-4}=\frac{1}{16} \), thus: \[ f(8\ln2)=16+256\left(\frac{1}{16}\right)=16+16=32. \] So the minimum point is \(\left(8\ln2,\,32\right)\). 6. **Graph Summary:** - **Domain:** All real numbers. - **Range:** \([32,\infty)\). - **\(x\)-intercepts:** None. - **\(y\)-intercept:** \((0,257)\). - **Critical point:** Global minimum at \(\left(8\ln2,\,32\right)\). - **Increasing/Decreasing Intervals:** - \( f(x) \) is decreasing on the interval \((-\infty,\,8\ln2)\). - \( f(x) \) is increasing on the interval \((8\ln2,\,\infty)\). 7. **Answering the Multiple Choice Selections:** - For the \(x\)-intercepts of \( f \), the correct choice is: **A. The \( x \)-intercept(s) of \( f \) is \( \) (Answer: There are no \( x \)-intercepts.) - For the \(y\)-intercept of \( f \), the correct choice is: **A. The \( y \)-intercept of \( f \) is \( y=257 \).** - For the increasing/decreasing behavior, the correct choice is: **A. The function \( f \) is increasing on the subinterval \( \left(8\ln2,\infty\right) \).** 8. **Sketch of the Graph:** - Plot the point \((0,257)\) for the \(y\)-intercept. - Indicate the global minimum at \(\left(8\ln2,\,32\right)\) (note that \( 8\ln2 \) is approximately \( 5.545 \)). - Show that as \( x \to -\infty \), \( f(x) \to \infty \) since the term \( 256e^{-0.5x} \) dominates. - Show that as \( x \to \infty \), \( f(x) \to \infty \) since the term \( e^{0.5x} \) dominates. - Draw a smooth U-shaped curve that is decreasing on \( (-\infty,\ 8\ln2) \) and increasing on \( (8\ln2,\ \infty) \). This completes the summary and the sketch with all the necessary information.