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

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Find the intervals in which $f(x)=\sin x-\cos x$ , where $0<x<2 \pi$ is increasing or decreasing.

Find the intervals in which $f(x)=\sin x-\cos x$ , where $0<x<2 \pi$ is increasing or decreasing.

Solution:

$\begin{array}{l} \text { Given } f(x)=\sin x-\cos x \\ \Rightarrow f^{\prime}(x)=\frac{d}{d x}(\sin x-\cos x) \\ \Rightarrow f^{\prime}(x)=\cos x+\sin x \end{array}$
For $f(x)$ let’s find critical point, we must have
$\begin{array}{l} \Rightarrow f^{\prime}(x)=0 \\ \Rightarrow \operatorname{Cos} x+\sin x=0 \\ \Rightarrow \operatorname{Tan}(x)=-1 \\ \Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4} \end{array}$
Here these points divide the angle range from 0 to $2 \pi$ as we have $x$ as angle
Clearly, $f^{\prime}(x)>0$ if $0<x<\frac{3 \pi}{4}$ and $\frac{7 \pi}{4}<x<2 \pi$ and $f^{\prime}(x)<0$ if $\frac{3 \pi}{4}<x<\frac{7 \pi}{4}$
Therefore, $f(x)$ 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 $\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$