What is the maximum value of the function $s i n x + c o s x$ ?

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Solution

Let $f (x) = s i n x + c o s x \Rightarrow f^{'} = c o s x - s i n x$
and $f "= - s i n x - c o s x = - (s i n x + c o s x)$
For maxima or minima put f'(x)=0
$\Rightarrow c o s x - s i n x = 0 \Rightarrow s i n x = c o s x \Rightarrow \frac{s i n x}{c o s x} = 1 \Rightarrow t a n x = 1 \Rightarrow x = \frac{π}{4}, \frac{5 π}{4} \dots$
Now, f"(x) will be negative when $(s i n x + c o s x)$ is positive i.e., when $s i n x a n d c o s x$ are both positive. Also, we know that sin x and cos x both are positive in the first quadrant.
Then, f"(x) will be negative when $x ϵ (0, \frac{π}{2})$
Thus, we consider $x = \frac{π}{4}$
$f " (\frac{π}{4}) = - (s i n \frac{π}{4} + c o s \frac{π}{4}) = - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = - \frac{2}{\sqrt{2}} = - \sqrt{2} < 0$
$∴$ By second derivative test. f will be maximum at $x = \frac{π}{4}$ and the maximum value of f is $f (\frac{π}{4}) = s i n \frac{π}{4} + c o s \frac{π}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

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