Question

Find intervals in which the function given by $f\left(x\right)=\sin 3x,x\in \left[0,\frac{\pi }{2}\right]$ is increasing function

Answer 1

Consider the given function ,

$f\left(x\right)=\sin 3x$ and $x\in [0,\frac{\pi }{2}]$

$f\left(x\right)=\sin 3x$ ….(1)

Differentiate with respect to x

$f^{{}^{\prime }}\left(x\right)=3\cos x$

For increasing and decreasing.

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

$3\cos 3x=0$

$\cos 3x=0$

When ,

$x=\frac{\pi }{2}\&\frac{3\pi }{2}$

$3x=\frac{\pi }{2}\&3x=\frac{3\pi }{2}$

$x=\frac{\pi }{6}\&x=\frac{\pi }{2}$

Since , $x=\frac{\pi }{6}\in [0,\frac{\pi }{2}]\&\frac{\pi }{2}x\in \left[0\frac{\pi }{2}\right]$

As show in fig. plotting the points

Since , $x\in [o,\frac{\pi }{2}]$ here we will start number line from 0 and end at $\frac{\pi }{2}$

Point $x=\frac{\pi }{6}$ divide the interval $[0,\frac{\pi }{2}]$ into two disjoint intervals

$[0,\frac{\pi }{6})$ and $(\frac{\pi }{6},\frac{\pi }{2}]$

Checking sign of $f^{{}^{\prime }}\left(x\right)$ =$3\cos x$

Case 1-

In $x\in (0,\frac{\pi }{6})$

$0<x<\frac{\pi }{6}$

$3\times 0<3x<\frac{3\pi }{6}$

$0<3x<\frac{3\pi }{6}$

When ,

$x\in (0,\frac{\pi }{6}),$ then 3$x\in (0,\frac{\pi }{2})$

And we know

$\cos 3\theta >0$ for $3x\in (0,\frac{\pi }{2})$

$\cos 3x>0,$ for $3x\in (0,\frac{\pi }{6})$

$3\cos 3x>0,$ for $3x\in (0,\frac{\pi }{6})$

At $x=0$

$f^{{}^{\prime }}\left(x\right)=3\cos 3\times 0=3$

At $x=\frac{\pi }{6}$

$$f^{{}^{\prime }}\left(x\right)=3\cos 3\times \left(\frac{\pi }{6}\right)\times 0=0$$

Since $f^{{}^{\prime }}\left(x\right)\ge 0$ for $x\in [0,\frac{\pi }{6}]$

Thus $f^{{}^{\prime }}\left(x\right)$ is increasing for $x\in [0,\frac{\pi }{6}]$