Question

The function $f\left(x\right)=\left(x^2-1\right)\left|x^2-3x+2\right|+\cos \left(\left|x\right|\right)$ is not differentiable at

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is D

$2$

Step 1: Find the critical point

$$\begin{gathered} f\left(x\right)=\left(x^2-1\right)\left|x^2-3x+2\right|+\cos \left(\left|x\right|\right) \\ \Rightarrow f\left(x\right)=g\left(x\right)+h\left(x\right) \end{gathered}$$

So, $h\left(x\right)=\cos \left(\left|x\right|\right)=\left\{\begin{array}{l}\cos \left(x\right),\mathrm{when}x\ge 0\\ -\cos \left(x\right),\mathrm{when}x<0\end{array}\right.$

Therefore it is differentiable at $x=0$

$\left|x^2-3x+2\right|=\left\{\begin{array}{l}\left(1-x\right)\left(2-x\right)\mathrm{if}\mathrm{x}<1\\ \left(x-1\right)\left(2-x\right)\mathrm{if}1\le x<2\\ \left(x-1\right)\left(x-2\right)\mathrm{if}\mathrm{x}\ge 2\end{array}\right.$

Therefore

$f\left(x\right)=\left\{\begin{array}{l}\left(x^2-1\right)\left(x-1\right)(x-1)+\cos \left(x\right)\mathrm{if}-\infty <\mathrm{x}<1\\ -\left(x^2-1\right)\left(x-1\right)(x-1)+\cos \left(x\right)\mathrm{if}1\le \mathrm{x}<2\\ \left(x^2-1\right)\left(x-1\right)(x-1)+\cos \left(x\right)\mathrm{if}2\le \mathrm{x}<\infty \end{array}\right.$

$x=1,2$ are critical point for differentiability

Because $f\left(x\right)$ is differentiable on other points in the domain

Step 2: Perform the differentiability test

Differentiability test at $x=1$

$$\begin{gathered} Lf\text{'}\left(1\right)=\underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ =\underset{x\to 1^{-}}{\lim }\left[\left(x^2-1\right)\left(x-2\right)+\frac{\cos \left(x\right)-\cos \left(1\right)}{x-1}\right] \\ =0-\sin \left(1\right) \\ Rf\text{'}\left(1\right)=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ =\underset{x\to 1^{+}}{\lim }\left[-\left(x^2-1\right)\left(x-2\right)+\frac{\cos \left(x\right)-\cos \left(1\right)}{x-1}\right] \\ =0-\sin \left(1\right) \\ \therefore Lf\text{'}\left(1\right)=Rf\text{'}\left(1\right) \end{gathered}$$

Differentiability test at $x=2$

$$\begin{gathered} Lf\text{'}\left(2\right)=\underset{x\to 2^{-}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2} \\ =\underset{x\to 2^{-}}{\lim }\left[-\left(x^2-1\right)\left(x-1\right)+\frac{\cos \left(x\right)-\cos \left(2\right)}{x-2}\right] \\ =-3-\sin \left(2\right) \\ Rf\text{'}\left(2\right)=\underset{x\to 2^{+}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2} \\ =\underset{x\to 2^{+}}{\lim }\left[\left(x^2-1\right)\left(x-1\right)+\frac{\cos \left(x\right)-\cos \left(2\right)}{x-2}\right] \\ =3-\sin \left(2\right) \\ \therefore Lf\text{'}\left(2\right)\ne Rf\text{'}\left(2\right) \end{gathered}$$

The function is not differentiable at $x=2$

Hence the correct option is $\left(\mathrm{D}\right)$.