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Find the intervals of concavity and the points of inflection of the following functions: $f(\theta)=\sin 2\theta $ in $ (0 , \pi )$

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  • Test for concavity Suppose $f$ is twice differentiable on an interval I.
  • (i) If $f''(x) > 0$ for all $x \in I$, then the graph of is concerve upward (convex downward ) on I.
  • (ii) If $ f''(x) < 0$ fro all $x \in I$. then the graph of f is conceve downward ( convex upward) on I.
$f(\theta)=\sin 2 \theta, \theta \in (0, \pi)$
Step 1:
$f'(\theta)= 2\cos 2 \theta$
$f''(x)= -4\sin 2 \theta$
Step 2:
$f''(\theta)=0=> -4 \sin 2 \theta=0$
$=>\sin 2 \theta =0 \;or \; 2 \theta =0, \pi ; 2 \pi$
$=> \theta =0, \large\frac{\pi}{2}$$,\pi.....$
The value $\theta= \large \frac{\pi}{2} $$\in (0,\pi)$
When $ 0 < \theta <\large\frac{\pi}{2}$$=> 0 <2 \pi <\pi$
$f''(\theta)=-4 \sin 2 \theta < 0 $ (Since $\sin 2 \theta > 0$) (concaved downward)
When $ \large\frac{\pi}{2}$$ < \theta < \pi =>\pi <2 \theta < \pi$
$f''(\theta)=-4 \sin 2 \theta > 0 $ (Since $\sin 2 \theta < 0$) (concaved upward)
Step 3:
$\therefore f''(0)$ changes from concave downwards to concave upward at
$\theta= \large\frac{\pi}{2}$
$\theta=\large\frac{\pi}{2}$ corresponds to a point of inflection $f(\large\frac{\pi}{2})$$=2 \sin \pi=0$
$\therefore$ the point of inflection is at $(\large\frac{\pi}{2}$$,0).$
answered Aug 5, 2013 by meena.p
 

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