logo

Ask Questions, Get Answers

X
 
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives

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: $ f (x) =\sin x-\cos x,0 < x <2\pi$

$\begin{array}{1 1} Maxima =\sqrt 3\; Minima = -\sqrt 2 \\ Maxima =\sqrt 2\; Minima = -\sqrt 2 \\ Maxima =\sqrt 2 Minima = \sqrt 2 \\ Maxima =\sqrt 2 Minima = -\sqrt 3 \end{array} $

Download clay6 mobile app

1 Answer

Toolbox:
  • $\large\frac{d}{dx}$$(\sin x)=\cos x$
  • $\large\frac{d}{dx}$$(\cos x)=-\sin x$
  • For Maxima and Minima $f'(x)=0$
Step 1:
$f(x)=\sin x-\cos x$
Differentiating with respect to x
$f'(x)=\cos x+\sin x$
$\qquad=\cos x[1+\large\frac{\sin x}{\cos x}]$
$\qquad=\cos x[1+\tan x]$
For Maxima and Minima
$f'(x)=0$
$\Rightarrow \cos x+\sin x=0$
$\cos x[1+\tan x]=0$
$1+\tan x=0$
$\tan x=-1$
$x=\large\frac{3\pi}{4},\frac{7\pi}{4}$
Step 2:
At $x=\large\frac{3\pi}{4}$
When $x$ is slightly $<\large\frac{3\pi}{4}$
$\cos x=-ve$
$\tan x=-(1+h)$
$\qquad=-1-h$
$1+\tan x=1+(-1-h)=-h$=-ve
$\cos x(1+\tan x)=(-)(-)=+ve$
Step 3:
when $x$ is slightly >$\large\frac{3\pi}{4}$
$\cos x=-ve$
$\tan x=-(1-h)$
$\qquad=-1+h$
$1+\tan x=-1-1+h=+h$=+ve
$f'(x)=\cos x(1+\tan x)$
$\qquad=(-)(+)=-ve$
$f'(x)$ changes sign from +ve to -ve.
Therefore At $x=\large\frac{3\pi}{4}$,there is a point of local maxima.
Step 4:
Local maximum value=$f(\large\frac{3\pi}{4})=$$\sin\large\frac{3\pi}{4}$$-\cos\large\frac{3\pi}{4}$
$\Rightarrow \large\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}$
$\Rightarrow \large\frac{2}{\sqrt 2}$
$\Rightarrow \sqrt 2$
Step 5:
At $x=\large\frac{7\pi}{4}$
Let $x$ is slightly < $\large\frac{7\pi}{4}$
$\cos x=+ve$
$\tan x=-(1+h)$
$1+\tan x=1-(1+h)=-h$=-ve
$f'(x)=\cos x(1+\tan x)$
$\qquad=(+)(-)=-ve$
When $x$ is slightly >$\large\frac{7\pi}{4}$
$\cos x=+ve$
$\tan x=-(1-h)=-1+h$
$1+\tan x=1-1+h=h=+ve$
$f'(x)=\cos x(1+\tan x)=(+)(+)=+ve$
$f'(x)$ changes sign from -ve to +ve .
Hence,there is a local minima at $x=\large\frac{7\pi}{4}$
Step 6:
Local minimum value=$f(\large\frac{7\pi}{4})$
$\qquad\qquad\qquad\quad\;\;=\sin x-\cos x$
$\qquad\qquad\qquad\quad\;\;=\sin \large\frac{7\pi}{4}$$-\cos\large\frac{7\pi}{4}$
$\qquad\qquad\qquad\quad\;\;=\large\frac{-1}{\sqrt 2}-\frac{1}{\sqrt 2}$
$\qquad\qquad\qquad\quad\;\;=\large\frac{-2}{\sqrt 2}$
$\qquad\qquad\qquad\quad\;\;=-\sqrt 2$
answered Aug 11, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

Related questions

Ask Question
Tag:MathPhyChemBioOther
...
X