Question

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

Answer 1

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

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

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

$\Rightarrow \cos 3x=0$

$\Rightarrow 3x=\frac{\pi }{2},\frac{3\pi }{2}$

$\Rightarrow x=\frac{\pi }{6}\in [0,\frac{\pi }{2}],\frac{\pi }{2}\in [0,\frac{\pi }{2}]$

$\therefore$ both values of $x$ are valid.

Since $x=\frac{\pi }{6}$ divide the interval $[0,\frac{\pi }{2}]$ in to two disjoint intervals $[0,\frac{\pi }{6})$ and $(\frac{\pi }{6},\frac{\pi }{2}]$

Case$:1$

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

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

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

$0<3x<\frac{\pi }{2}$

So, when $x\in (0,\frac{\pi }{6})\Rightarrow 3x\in (0,\frac{\pi }{2})$

And we know that

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

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

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

$f^{\prime }\left(x\right)>0$ for $x\in (0,\frac{\pi }{6})$

At $x=0,f^{\prime }\left(x\right)=3\cos 0=3\times 1=3$

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

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

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