How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$? How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$ ?
Please let me know the fastest method you know of for such type of problems. Is there any way other than finding the left hand and right hand derivative using the concept of limits (That makes it huge)? Can intuition be used in any way?
 A: $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\\$ then
$$f(x)=\sqrt{(\sqrt{x-2}+\sqrt{2})^2}+\sqrt{(\sqrt{x-2}-\sqrt{2})^2}=\\|\sqrt{x-2}+\sqrt{2}|+|\sqrt{x-2}-\sqrt{2}|$$ now note that $ x\geq 2$   so ,when $$x \rightarrow 4^+$$
$$f(x)=\sqrt{x-2}+\sqrt{2} +\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\\f'_{4^+}=2\frac{1}{2\sqrt{x-2}}=\frac{1}{\sqrt{2}}$$
when $$x \rightarrow 4^-$$
$$f(x)=\sqrt{x-2}+\sqrt{2} -(\sqrt{x-2}-\sqrt{2})=2\sqrt{2}\\f'_{4^-}=0$$
so it is not differentiable at $x=4$ 
A: Observe that $(f(x))^2$ is given by
$$(x + 2\sqrt{2x - 4}) + 2 \sqrt{x + 2\sqrt{2x - 4}}\sqrt{x - 2\sqrt{2x - 4}} + (x - 2\sqrt{2x - 4})$$
$$= 2x + 2\sqrt{x^2 - 4(2x - 4)}$$
$$= 2x + 2\sqrt{(x - 4)^2}$$
$$= 2x + 2|x - 4|$$
Since $f(x) \geq 0$, for $x$ near $4$ we therefore have
$$ f(x) = \sqrt{2x + 2|x - 4|}$$
For $x > 4$ this is $\sqrt{4x - 8}$ and for $x < 4$ this is just $\sqrt{8}$. Hence the left and right derivatives do not match at $x = 4$ and the function is not differentiable there.
A: Expand $x-2\sqrt{2x-4}$ at $x=4$ to get $\frac{1}{8}(x-4)^2+O((x-4)^3)$. Hence
$$
\sqrt{x-2\sqrt{2x-4}} \approx \sqrt{\frac{1}{8}}|x-4|,
$$
and this should make it clear why $f$ is not differentiable at $x=4$.
