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Prove that the following functions are not monotonic in the intervals given. $x \sin x$ on $[0 , \pi ]$

Note: This is part 3rd of a 4 part question, split as 4 separate questions here.

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  • If $f(x)$ is defined on an open interval $I$ and $f'(x)$ takes both positive and negative values on $I$, then $f$ is not monotonic on $I$
$f(x)= x \sin x \;on \;[0,\pi]$
Step 1:
$f'(x)= x \cos x +\sin x$
$f'(0)=0$
$f'(\pi)=-\pi$
Step 2:
$f'(x) >0$ in $\bigg(0, \large\frac{\pi}{2}\bigg)$$ \qquad (since \; \cos x, \sin x,x >0)$
In $\bigg(\large\frac{\pi}{2}$$,\pi\bigg)$ consider $x=\large\frac{3\pi}{4}$
Step 3:
$f' \bigg( \large\frac{3 \pi}{4}\bigg)=\large\frac{3 \pi}{4} $$\cos \large\frac{3 \pi}{4}$$+\sin \large\frac{3\pi}{4}$
$\qquad\qquad=\bigg[\large\frac{3 \pi}{4}\bigg( \large\frac{-1}{\sqrt 2} \bigg)+\frac{1}{\sqrt 2}\bigg]$
$\qquad\qquad=\large\frac{1}{\sqrt 2} \bigg[1- \large \frac{3 \pi}{4}\bigg] $$<0$ since $\large\frac{3 \pi}{4}$$ >1$
$\therefore f'(x)$ takes positive and -negative values in $ [0,\pi]$
It is not monotonic in $[0, \pi]$

 

answered Jul 30, 2013 by meena.p
 

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