Question

The set of points where f(x) = cos |x| is differentiable, is ____________.

Answer 1

We know

$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right.$

$\therefore f\left(x\right)=\cos \left|x\right|=\left\{\begin{array}{ll}\cos x,& x\ge 0\\ \cos \left(-x\right),& x<0\end{array}\right.$

$\Rightarrow f\left(x\right)=\cos \left|x\right|=\left\{\begin{array}{ll}\cos x,& x\ge 0\\ \cos x,& x<0\end{array}\right.\left[\cos \left(-\theta \right)=\cos \theta \right]$

We know that, cosine function is differentiable in its domain. So, f(x) is differentiable for all x < 0 and x > 0.

Let us check the differentiability of $f\left(x\right)=\cos \left|x\right|$ at x = 0.

Now,

$Lf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{-h}$

$\Rightarrow Lf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{\cos \left(-h\right)-\cos 0}{-h}$

$\Rightarrow Lf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{\cos h-1}{-h}$

$\Rightarrow Lf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{-h}$

$\Rightarrow Lf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\times \underset{h\to 0}{\lim }\sin \frac{h}{2}$

$\Rightarrow Lf\text{'}\left(0\right)=1\times 0$

$\Rightarrow Lf\text{'}\left(0\right)=0$

And

$Rf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$

$\Rightarrow Rf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{\cos h-\cos 0}{h}$

$\Rightarrow Rf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{\cos h-1}{h}$

$\Rightarrow Rf\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{h}$

$\Rightarrow Rf\text{'}\left(0\right)=-\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\times \underset{h\to 0}{\lim }\sin \frac{h}{2}$

$\Rightarrow Rf\text{'}\left(0\right)=-1\times 0$

$\Rightarrow Rf\text{'}\left(0\right)=0$

$\therefore Lf\text{'}\left(0\right)=Rf\text{'}\left(0\right)$

So, f(x) is differentiable at x = 0. Thus, the function f(x) is differentiable everywhere.

Hence, the set of points where $f\left(x\right)=\cos \left|x\right|$ is differentiable is R (set of real real numbers).


The set of points where f(x) = cos |x| is differentiable, is _____R_____.