Question

The maximum value of $f(x) = e^{\sin x} + e^{\cos x}$, where $x \in \mathbb{R}$, is

• A. 2e
• B. 2 √e
• C. 2e 1/√2
• D. 2e-1/√2

Answer 1

The Correct Option is C

We are tasked with finding the maximum value of the function: \[ f(x) = e^{\sin x} + e^{\cos x} \] where \( x \in \mathbb{R} \).

Step 1: Analyzing the function
The function \( f(x) \) is the sum of two exponential functions, one in terms of \( \sin x \) and the other in terms of \( \cos x \). To find the maximum value of \( f(x) \), we need to analyze the behavior of these terms. Both \( \sin x \) and \( \cos x \) are bounded by \( -1 \) and \( 1 \), which means \( e^{\sin x} \) and \( e^{\cos x} \) will be bounded by \( e^{-1} \) and \( e^{1} \), respectively. 

Step 2: Maximize the sum
To maximize \( f(x) = e^{\sin x} + e^{\cos x} \), we look for the values of \( x \) where \( \sin x \) and \( \cos x \) are maximized simultaneously. The values of \( \sin x \) and \( \cos x \) both reach their maximum at \( x = \frac{\pi}{4} \), where: \[ \sin \left( \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \] Thus, at \( x = \frac{\pi}{4} \), we have: \[ f\left( \frac{\pi}{4} \right) = e^{\frac{1}{\sqrt{2}}} + e^{\frac{1}{\sqrt{2}}} = 2 e^{\frac{1}{\sqrt{2}}} \] 

Step 3: Conclusion
Thus, the maximum value of \( f(x) \) is: \[ \boxed{2e^{1/\sqrt{2}}} \]

Answer:

\[ \boxed{2e^{1/\sqrt{2}}} \]