# Find the intervals in which

Question:

Find the intervals in which $f(x)=\sin x-\cos x$, where $0 Solution: Given:- Function$f(x)=\sin x-\cos x, 0

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

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

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

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

For $f(x)$ lets find critical point, we must have

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

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

$\Rightarrow \tan (x)=-1$

$\Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4}$

clearly, $f^{\prime}(x)>0$ if $0 and$f^{\prime}(x)<0$if$\frac{3 \pi}{4}

Thus, $f(x)$ increases on $\left(0, \frac{3 \pi}{4}\right) \cup\left(\frac{7 \pi}{4}, 2 \pi\right)$

and $f(x)$ is decreasing on interval $\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$