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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the local maximum and local minimum value of the function. \[f(x)=\sin x+\frac{1}{2}\cos 2x\;in\;[0,\pi/2]\]

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Toolbox:
  • For local maximum or minimum $ f '(x)=0$
  • For maximum $ f "(x)$ is negative and for minimum $ f "(x)$ is positive.
Step 1
$ f'(x)=cos\: x-sin2x$
$ f'(x)=0$ at $ x = \large\frac{\pi}{6}, \large\frac{\pi}{2}$
Step 2
$ f"(x)= -sin\: x - 2cos\:2x$
is negative at $ x = \large\frac{\pi}{6}$ and is positive at $ \large\frac{\pi}{2}$
Step 3
At $ x = \large\frac{\pi}{6}\: f(x)$ has got maxima and maximum value is $f\bigg( \large\frac{\pi}{6} \bigg)=\large\frac{3}{4}$
At $ x =\large\frac{\pi}{2}\: f(x)$ has got minima and minimum value is $ f \bigg( \large\frac{\pi}{2} \bigg) = \large\frac{1}{2}$
answered Apr 12, 2013 by thanvigandhi_1
 

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