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Find the absolute maximum and absolute minimum values of $f$ on the given interval: $\;f(x)=x-2\cos x , [-\pi ,\pi ]$

Note: This is part 7th of a 7 part question, split as 7 separate questions here.

1 Answer

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  • To find the absolute maximum and minimum values of a continuous function f on a closed interval $[a,b]$
  • (i) Find the values of the critical numbers of f in $(a,b)$
  • (ii) Find the value of $f(a)$ and $f(b)$
  • (iii) The largest of the values from (i) and (ii) is the absolute maximun value, the smallest of these values is the absolute minimum value.
$f(x)=x-2 \cos x$ is continous on $[-\pi,\pi]$
$f'(x)=1+2 \sin x$
$f'(x)=0=>\sin x =-\large\frac{1}{2}$ or $x=-\pi +\large\frac{\pi}{6}, \large\frac{-\pi}{6}$
$ie \;x= \large\frac{-2 \pi}{6},\frac{-\pi}{6}$ ie $ \large\frac{-\pi}{3},\frac{-\pi}{6}$
Step 2:
The critical values are
$f(\large\frac{-\pi}{3})=\frac{-\pi}{3}$$-2 (\large\frac{1}{2})=\large\frac{-\pi}{3}$$-1$ (appr -2.05)
$f(\large\frac{-\pi}{6})=\frac{-\pi}{6}$$-2 (\large\frac{\sqrt 3}{2})=\large\frac{-\pi}{6}$$-\sqrt 3$ (appr -2.25)
The values of $f$ at the end points are
$f(-\pi)=-\pi-2(-1)=2 -\pi$ (appr -1.14)
$f(\pi)=\pi-2(-1)=2 +\pi$ (appr 5.14)
Step 3:
Comparing the values,
The absolute maximum is $f(\pi)=2 +\pi$
The absolute minimum is $f(-\large\frac{\pi}{6})=\frac{-\pi}{6}$$-\sqrt 3$
answered Jul 31, 2013 by meena.p

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