# Find the local maximum and minimum values of the following: $\sin^{2} \theta , [0 ,\pi ]$

Note: This is part 5th of a 6 part question, split as 6 separate questions here.

## 1 Answer

Toolbox:
• Second derivative test: Suppose $f$ is continuous on an open interval that contains
• (i)If $f'(c)=0\; and\; f''(c) >0$ then $f$ has a local minimum at $c$.
• (ii)If $f'(c)=0\;and \; f''(c)<0$ then $f$ has a local maximum at $c$
$f(\theta)=\sin ^2 \theta \;in\; [\theta,\pi]$
Step 2:
$f'(\theta)=2 \sin \theta \cos \theta=\sin 2 \theta$
$f'(\theta)=0$ at the turning points
$f'(\theta)=0=>\sin 2 \theta=0 \qquad 2 \theta=n \pi \quad n \in Z$
$=>\theta=\large\frac{\pi}{2} $$n \in z of these \theta=0, \large\frac{\pi}{2},$$ \pi \in\quad$$[0,\pi] The turning points are in the interval (0,\pi) Step 2: When \theta=\large\frac{\pi}{2}$$ f(\theta)=1$
$\therefore$ there is a turning point at $(\large\frac{\pi}{2},$$1) Step 3: f''(\theta)=2 \cos \theta When \theta=\large\frac{\pi}{2}$$ f''(\theta)=-2 < 0$
there is a local maximum at $\theta=\large\frac{\pi}{2}$
Step 4:
$\therefore f(\theta)$ has a local maximum at $\bigg( \large\frac{\pi}{2},1\bigg)$

answered Aug 1, 2013 by
edited Aug 1, 2013 by meena.p

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