Question

The set of points where the function $f\left(x\right)=\left\{\begin{array}{cc}x+1,& x<2\\ 2x-1,& x\ge 2\end{array}\right.$is not differentiable, is ____________.

Answer 1

The given function is $f\left(x\right)=\left\{\begin{array}{cc}x+1,& x<2\\ 2x-1,& x\ge 2\end{array}\right.$.

(x + 1) and (2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x < 2 and for all x > 2.

So, we need to check the differentiability of f(x) at x = 2.

We have

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

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

$\Rightarrow Lf\text{'}\left(2\right)=\underset{h\to 0}{\lim }\frac{3-h-3}{-h}$

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

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

And

$Rf\text{'}\left(2\right)=\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}$

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

$\Rightarrow Rf\text{'}\left(2\right)=\underset{h\to 0}{\lim }\frac{3+2h-3}{h}$

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

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

$\therefore Lf\text{'}\left(2\right)\ne Rf\text{'}\left(2\right)$

So, f(x) is not differentiable at x = 2.

Thus, the set of points where the function f(x) is not differentiable is {2}.


The set of points where the function $f\left(x\right)=\left\{\begin{array}{cc}x+1,& x<2\\ 2x-1,& x\ge 2\end{array}\right.$is not differentiable, is _____{2}______.