Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

Sketch the curves $y = |\sin x| + |\cos x|$ .

Accepted Answer

Given: $y = |\sin x| + |\cos x| = f (x)$

Period of $|\sin x|$ and $|\cos x|$ is $π$

For checking period of $f (x)$ , check for $T = \frac{π}{2}$ also

$f (\frac{π}{2} + x) = |\sin (\frac{π}{2} + x)| + |\cos (\frac{π}{2} + x)|$

$\Rightarrow f (\frac{π}{2} + x) = |\cos x| + |- \sin x|$

$\Rightarrow f (\frac{π}{2} + x) = |\cos x| + |\sin x| = f (x)$

So, $f (x)$ is periodic with period $T = \frac{π}{2}$

Replace $x \to - x,$ in $f (x)$

for checking even/odd

$f (- x) = |\sin (- x)| + |\cos (- x)|$

$\Rightarrow f (- x) = |\sin x| + |\cos x| = f (x)$

So, $f (x)$ is an even function graph of $f (x)$ is symmetric with respect to $x$ -axis.

As period $= \frac{π}{2},$ so consider

$x \in [0, \frac{π}{2}]$

$f (x) = \sin x + \cos x$

$∵$ For $x \in [0, \frac{π}{2}],$

$\sin x \geq 0, \cos x \geq 0$

$y = \sin x + \cos x$

$\Rightarrow y = \sqrt{2} (\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x)$

$\Rightarrow y = \sqrt{2} \sin (x + \frac{π}{4})$

So, for $x \in [0, \frac{π}{2}]$

So, the graph of $y = |\sin x| + |\cos x|$

Sketch the curves $y = |\sin x| + |\cos x|$ .

Important Questions on Curvature and Transformations

Learn and Prepare for any exam you want !

Sketch the curves y=sinx+cosx.

Important Questions on Curvature and Transformations

Learn and Prepare for any exam you want !

Sketch the curves $y = |\sin x| + |\cos x|$ .