Question

An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is ____________.

Answer 1

Consider the function $g\left(x\right)=\left|x-1\right|+\left|x+1\right|$.

$\left|x-1\right|=\left\{\begin{array}{ll}x-1,& x\ge 1\\ -\left(x-1\right),& x<1\end{array}\right.$

$\left|x+1\right|=\left\{\begin{array}{ll}x+1,& x\ge -1\\ -\left(x+1\right),& x<-1\end{array}\right.$

$\therefore g\left(x\right)=\left|x-1\right|+\left|x+1\right|=\left\{\begin{array}{ll}-\left(x-1\right)-\left(x+1\right),& x<-1\\ -\left(x-1\right)+x+1,& -1\le x<1\\ x-1+x+1,& x\ge 1\end{array}\right.$

$\Rightarrow g\left(x\right)=\left|x-1\right|+\left|x+1\right|=\left\{\begin{array}{ll}-2x,& x<-1\\ 2,& -1\le x<1\\ 2x,& x\ge 1\end{array}\right.$

When x < −1, g(x) = −2x which being a polynomial function is continuous and differentiable.

When −1 ≤ x < 1, g(x) = 2 which being a constant function is continuous and differentiable.

When x ≥ 1, g(x) = 2x which being a polynomial function is continuous and differentiable.

Let us check the continuity and differentiability of g(x) at x = −1 and x = 1.

At x = −1,

LHL = $\underset{x\to -1^{-}}{\lim }g\left(x\right)=\underset{x\to -1}{\lim }\left(-2x\right)=-2\times \left(-1\right)=2$

RHL = $\underset{x\to -1^{+}}{\lim }g\left(x\right)=\underset{x\to -1}{\lim }2=2$

$g\left(-1\right)=2$

Since $\underset{x\to -1^{-}}{\lim }g\left(x\right)=\underset{x\to -1^{+}}{\lim }g\left(x\right)=g\left(-1\right)$, so the function g(x) is continuous at x = −1.

At x = 1,

LHL = $\underset{x\to 1^{-}}{\lim }g\left(x\right)=\underset{x\to 1}{\lim }2=2$

RHL = $\underset{x\to 1^{+}}{\lim }g\left(x\right)=\underset{x\to 1}{\lim }2x=2\times 1=2$

$g\left(1\right)=2\times 1=2$

Since $\underset{x\to 1^{-}}{\lim }g\left(x\right)=\underset{x\to 1^{+}}{\lim }g\left(x\right)=g\left(1\right)$, so the function g(x) is continuous at x = 1.

Thus, the function g(x) is continuous everywhere i.e. for all x ∈ R.

Now,

$Lg\text{'}\left(-1\right)=\underset{h\to 0}{\lim }\frac{g\left(-1-h\right)-g\left(-1\right)}{-h}$

$\Rightarrow Lg\text{'}\left(-1\right)=\underset{h\to 0}{\lim }\frac{-2\left(-1-h\right)-2}{-h}$

$\Rightarrow Lg\text{'}\left(-1\right)=\underset{h\to 0}{\lim }\frac{2h}{-h}$

$\Rightarrow Lg\text{'}\left(-1\right)=-2$

And

$Rg\text{'}\left(-1\right)=\underset{h\to 0}{\lim }\frac{g\left(-1+h\right)-g\left(-1\right)}{h}$

$\Rightarrow Rg\text{'}\left(-1\right)=\underset{h\to 0}{\lim }\frac{2-2}{-h}$

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

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

$\therefore Lg\text{'}\left(-1\right)\ne Rg\text{'}\left(-1\right)$

So, g(x) is not differentiable at x = −1.

Also,

$Lg\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{g\left(1-h\right)-g\left(1\right)}{-h}$

$\Rightarrow Lg\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{2-2\times 1}{-h}$

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

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

And

$Rg\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{g\left(1+h\right)-g\left(1\right)}{h}$

$\Rightarrow Rg\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{2\left(1+h\right)-2\times 1}{h}$

$\Rightarrow Rg\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{2h}{h}$

$\Rightarrow Rg\text{'}\left(1\right)=2$

$\therefore Lg\text{'}\left(1\right)\ne Rg\text{'}\left(1\right)$

So, g(x) is not differentiable at x = 1.

Thus, the function g(x) is differentiable everywhere except at x = −1 and x = 1.


An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is $\underline{g\left(x\right)=\left|x-1\right|+\left|x+1\right|}$.