Question

Prove that the greatest integer function is defined by,
$$f(x)=[x],0<x<3$$ is not differentiable at $$x=1$$ and $$x=\text{ 2}$$.

Answer 1

Hint: In this question first we will check the continuity of the given function at a given point, if it is discontinuous at that given point, it will also be non-differentiable at that point.

Complete step-by-step answer:
As we are given with a function,
$$\Rightarrow f(x)=[x],0<x<3$$
$$\Rightarrow$$As, we know that, if $$f(x)$$ is not continuous at a point,
Then it will not be differentiable at that point too.
$$\Rightarrow$$So, let us check for continuity of $$f(x)=\left[x\right]$$ at $$x=1$$ and $$x=\text{ 2}$$.
Checking continuity at $$x=1$$
$$\Rightarrow \underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}[x]=0$$
$$\Rightarrow$$And, $$\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}[x]=1$$
As, we have seen above that $$\underset{x\to 1^{-}}{lim}f(x)\ne \underset{x\to 1^{+}}{lim}f(x)$$
$$\Rightarrow$$Therefore, $$f(x)$$ is neither continuous nor differentiable at $$x=1$$.
Now, checking continuity at $$\text{x}=2$$.
$$\Rightarrow \underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}[x]=1$$
$$\Rightarrow$$And, $$\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}[x]=2$$
As, we have seen above that, $$\underset{x\to 2^{-}}{lim}f(x)\ne \underset{x\to 2^{+}}{lim}f(x)$$.
$$\Rightarrow$$Therefore, $$f(x)$$ is neither continuous nor differentiable at $$\text{x}=2$$.
$$\Rightarrow$$Hence, $$f(x)$$ is neither differentiable at $$x=1$$
  nor differentiable at $$\text{x}=2$$.

Note: Whenever we come up with this type of problem we are asked to check whether the function, $$f(x)$$ is differentiable or not. Then first we should check the continuity of the given function, if it is continuous then we have to check whether the function is differentiable or not.