# The set of points where the function

Question:

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

Solution:

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

$(x+1)$ and $(2 x-1)$ are polynomial functions which are differentiable at each $x \in \mathrm{R} .$ 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

$L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$

$\Rightarrow L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{(2-h+1)-(2 \times 2-1)}{-h}$

$\Rightarrow L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{3-h-3}{-h}$

$\Rightarrow L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{-h}{-h}$

$\Rightarrow L f^{\prime}(2)=1$

And

$R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$\Rightarrow R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{[2(2+h)-1]-(2 \times 2-1)}{h}$

$\Rightarrow R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{3+2 h-3}{h}$

$\Rightarrow R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{2 h}{h}$

$\Rightarrow R f^{\prime}(2)=2$

$\therefore L f^{\prime}(2) \neq R f^{\prime}(2)$

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 $f(x)=\left\{\begin{array}{cl}x+1, & x<2 \\ 2 x-1, & x \geq 2\end{array}\right.$ is not differentiable, is ___{2}______