Question

Find the intervals in which the function $f$ given by $f\left(x\right)=\sin x+\cos x,0\le x\le 2\pi$ is increasing or decreasing.

Answer 1

$f\left(x\right)=\sin x+\cos x,0\le x\le 2\pi$
Differentiating w.r.t. $x$, we get
$f^{\prime }\left(x\right)=\cos x-\sin x$
Now, $f^{\prime }\left(x\right)=0$
$\Rightarrow \cos x-\sin x=0$
$\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4}$
Plotting points on number line


$f^{\prime }\left(x\right)=\cos x-\sin x$
When $x\in [0,\frac{\pi }{4})$,
$\cos x-\sin x>0$
$\Rightarrow f^{\prime }\left(x\right)>0$
So, $f\left(x\right)$ is increasing
When $x\in (\frac{\pi }{4},\frac{5\pi }{4})$,
$\cos x-\sin x<0$
$\Rightarrow f^{\prime }\left(x\right)<0$
So, $f\left(x\right)$ is decreasing.
When $x\in (\frac{5\pi }{4},2\pi ]$,
$\cos x-\sin x>0$
$\Rightarrow f^{\prime }\left(x\right)>0$
So, $f\left(x\right)$ is increasing.
Thus, $f\left(x\right)$ is increasing for
$x\in [0,\frac{\pi }{4})\cup (\frac{5\pi }{4},2\pi ]$
$f\left(x\right)$ is decreasing for
$x\in (\frac{\pi }{4},\frac{5\pi }{4})$