# How to prove the given function is not differentiable analytically?

Well the question presented to me is this. The given function is,
$$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{2}x + 2,\,\;\;\;x < 2}\\{\sqrt {2x} ,\;\;\;\;\;\;x \ge 2}\end{array}} \right.$$ Now have to check whether the given function is differenciable at $x=2$ ?

My approach:
For this function to be differentiable, the left hand derivative and the right hand derivative must exits, and both be equal.
Left hand derivative:
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ - }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left( {\frac{1}{2}x + 2} \right) - \left( {\frac{1}{2}\left( 2 \right) + 2} \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left( {\frac{1}{2}x + 2} \right) - 3}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\frac{1}{2}x - 1}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x - 2}}{{2\left( {x - 2} \right)}}\end{array}$$
This gives,
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ - }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{}}{{2}}\\ = \frac{1}{2}\end{array}$$

Right hand derivative:
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ + }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\left( {\sqrt {2x} } \right) - \left( {\sqrt {2\left( 2 \right)} } \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - \sqrt 4 }}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - 2}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {2x} - 2}}{{x - 2}} \times \frac{{\sqrt {2x} + 2}}{{\sqrt {2x} + 2}}\end{array}$$
This gives,
$$\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ + }} \frac{{f\left( x \right) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{2x - 4}}{{\left( {x - 2} \right)\left( {\sqrt {2x} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{2}{{\left( {\sqrt {2x} + 2} \right)}}\\ = \frac{2}{{2 + 2}}\\ = \frac{1}{2}\end{array}$$

Problem is right hand derivative and left hand derivative are coming same. Fooling me that its differentiable at $x=2$.

The graph of the function is,
$f(x)$ Graph
The function is discontinues at $x=2$. So, its not differentiable at $x=2$.

Where have I gone wrong ? How do I prove analytically without taking help of graph that the function is not differentiable at $x=2$?

• You have made a mistake in value of $f(2)$. It is given by $f(2)=2$ and not $3$ as you have tried to do while calculating left hand derivative. The left hand derivative is $-\infty$. Also note that the analytical thing is the real stuff and graphical arguments are only an approximation to the analytical argument and are meant to help in understanding the concepts. May 25 '17 at 14:27
• @ParamanandSingh I think it would be correct to take this as function $\frac{1}{2}x + 2$ when approaching from left.
– user449525
May 25 '17 at 14:45
• The value $f(2)$ does not depend on left right. It depends on point $2$. Do not confuse value of a function with left or right limit of the function. These are different concepts and value of a function is a much simpler concept than the limit of a function. May 25 '17 at 14:45
• Yes.. Got the point. I did a mistake there. Thanks.
– user449525
May 25 '17 at 14:47
• Do you seriously need help in getting values of $f$? Check definition of $f$. If $x<2$ the value $f(x)$ is given by formula $(x/2)+2$ so $f(1)=5/2$. But if $x\geq 2$ then $f(x)$ is given by formula $\sqrt{2x}$ and hence $f(2)=\sqrt{4}=2$. I hope this is clear. May 25 '17 at 14:51

$$f (2^-)=\lim_{x\to 2^-} (\frac {x}{2}+2)=1+2=3$$ $$f (2^+)=\lim_{2^+}\sqrt {2x}=2$$

$f$ is not continuous at $x=2$ thus it is not differentiable at $x=2$.

By definition, differentiable at $x=x_0$ means $$\exists L\in \mathbb R \;\exists \eta>0 :\forall x\in (2-\eta,2+\eta)$$

$$f (x)=f (2)+(x-2)\Bigl (L+\epsilon (x)\Bigr)$$

with $$\lim_{x\to 2}\epsilon (x)=0$$

• Have you arrived at $$f (2^-)=1+2=3\neq f (2^+)=2$$ without using help of graph ? If yes, please do explain a bit.
– user449525
May 25 '17 at 12:08
• @Mathanatic Look now May 25 '17 at 12:12
• What's wrong in what I did? Is that definition incorrect? + How $$f (2^-)=1+2=3\neq f (2^+)=2$$ came ?
– user449525
May 25 '17 at 12:17
• That doesn't explain why the OP's work is incorrect though.
– user370967
May 25 '17 at 13:35