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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: $ (iii)\: h(x)=\sin x+\cos x$

This is third part of multipart q3

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  • $f$ is said to have a maximum value in $I$ , if there exist a point c in I such that $f(c) \geq f (x)$ for all $x \in I$.The number $f( c)$ is called the maximum value of f in I and the point c is called a point of maximum value of f in I
  • $f$ is said to have a minimum value in $I$ , if there exist a point $c$ in I such that $f(c) \leq f (x)$ for all $x \in I$.The number $f(c)$ is called the minimum value of f in I and the point $c$ in this case is called a point of minimum value of $f$ in I
  • $f$ is said to have a extreme value in $I$ , if there exist a point $c$ in I such that f(c) is either a maximum value or minimum value of $f$ in $I$. The number $f (c)$ in this case is called the extreme value of $f$ in $I$ and the point $c$ is called the extreme point.
Step 1:
$h(x)=\sin x+\cos x$
Differentiating with respect to x
$h'(x)=\cos x-\sin x$
$\qquad=\cos x[1-\large\frac{\sin x}{\cos x}]$
$\qquad=\cos x[1-\tan x]$
For maxima and minima
$h'(x)=0$
$\Rightarrow \cos x-\sin x=0$
(or)$\tan x=1$
$x=\tan^{-1}1$
$x=\large\frac{\pi}{4}$
$x=\large\frac{\pi}{4}$$\in [0,\large\frac{\pi}{2}]$
Step 2:
At $x=\large\frac{\pi}{4}$
$h'(x)=\cos x[1-\tan x]$
When $x$ is slightly < $\large\frac{\pi}{4}$
$\cos x=+ve$
$\tan x=1-h$
Where $h$ is small.
$1-\tan x=1-(1-h)=+ve$
$h'(x)=\cos x(1-\tan x)$
$\qquad=(+)(+)=+ve$
Step 3:
When $x$ is slightly > $\large\frac{\pi}{4}$
$\cos x=+ve$
$\tan x=1+h$
Where $h$ is small.
$1-\tan x=1-(1-h)$
$\qquad\;\;\;\;\;\;=-h=-ve$
Step 4:
Hence there is a local maxima at $x=\large\frac{\pi}{4}$
Local maximum value =$h\big(\large\frac{\pi}{4}\big)$$=\sin\large\frac{\pi}{4}+$$\cos \large\frac{\pi}{4}$
$\qquad\qquad\qquad\qquad\qquad\quad=\large\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}$
$\qquad\qquad\qquad\qquad\qquad\quad=\large\frac{2}{\sqrt 2}$$=\sqrt 2$
answered Aug 7, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

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