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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?

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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.

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$f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\\$ 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 $x=4$

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  • $\begingroup$ How does this relate with differentiability at x=4? $\endgroup$ – user220382 Jul 28 '15 at 15:58
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    $\begingroup$ $\sqrt{x-2}-\sqrt{2}<0$ if $x<4$ $\endgroup$ – Empy2 Jul 28 '15 at 16:00
  • $\begingroup$ The answer is a combination of this and @Michael. When $x>4$, $f(x)=2\sqrt{x-2}$. When $x<4$, $f(x)=2\sqrt{2}$. Then use this when computing the left/right limits. $\endgroup$ – Moya Jul 28 '15 at 16:03
  • $\begingroup$ All observations up to the sum of absolute values are good. Then one can use linearity of derivative together with derivative of each term. $\endgroup$ – mathreadler Jul 28 '15 at 16:04
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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 $x=4$.

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