The maximum value of f(x) = e^{sin x} + e^{co | vedclass

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(C) Given $f(x) = e^{\sin x} + e^{\cos x}$.To find the maximum value, we use the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality or derivative a

Vedclass · Accepted Answer

$2e^{1/\sqrt{2}}$

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(C) Given $f(x) = e^{\sin x} + e^{\cos x}$.To find the maximum value, we use the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality or derivative analysis.Let $u = \sin x$ and $v = \cos x$. We know that $u^2 + v^2 = 1$.Using the property that for $f(x) = e^{\sin x} + e^{\cos x}$, the function attains its maximum when $\sin x = \cos x$.Setting $\sin x = \cos x$, we get $	an x = 1$, which implies $x = \frac{\pi}{4} + n\pi$.For $x = \frac{\pi}{4}$, $\sin x = \frac{1}{\sqrt{2}}$ and $\cos x = \frac{1}{\sqrt{2}}$.Substituting these values into $f(x)$:$f\left(\frac{\pi}{4}ight) = e^{1/\sqrt{2}} + e^{1/\sqrt{2}} = 2e^{1/\sqrt{2}}$.Thus, the maximum value is $2e^{1/\sqrt{2}}$.

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

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