The maximum value

Question:

The maximum value of $f(x)=\sin x+\cos x$ is _______________.

Solution:

The given function is $f(x)=\sin x+\cos x$.

$f(x)=\sin x+\cos x$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=\cos x-\sin x$

For maxima or minima,

$f^{\prime}(x)=0$

$\Rightarrow \cos x-\sin x=0$

$\Rightarrow \cos x=\sin x$

 

$\Rightarrow \tan x=1$

$\Rightarrow x=\frac{\pi}{4}, \frac{5 \pi}{4}$   (Let us only consider the values of $x \in[0,2 \pi]$ )

Now,

$f^{\prime \prime}(x)=-\sin x-\cos x$

At $x=\frac{5 \pi}{4}$, we have

$f^{\prime \prime}\left(\frac{5 \pi}{4}\right)=-\sin \frac{5 \pi}{4}-\cos \frac{5 \pi}{4}=-\left(-\frac{1}{\sqrt{2}}\right)-\left(-\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}>0$

So, $x=\frac{5 \pi}{4}$ is the point of local minimum of $f(x)$.

At $x=\frac{\pi}{4}$, we have

$f^{\prime \prime}\left(\frac{\pi}{4}\right)=-\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0$

So, $x=\frac{\pi}{4}$ is the point of local maximum of $f(x)$.

$\therefore$ Maximum value of $f(x)=f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Thus, the maximum value of $f(x)=\sin x+\cos x$ is $\sqrt{2}$.

The maximum value of $f(x)=\sin x+\cos x$ is $\sqrt{2}$

 

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