Suppose a real valued function $f:\mathbb R \to \mathbb R$ is continuous everywhere. Is it possible to construct $f$ that is differentiable at only one point? If possible give an example also.
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It is possible. Start with a function $g$ that is contiuous everywhere and nowhere differentiable, like the Weierstraß function.
Then choose the point where you want it differentiable, say $a$, and set
$$f(x) = (x-a)\cdot g(x).$$
Then $f$ is differentiable in $a$, with $f'(a) = g(a)$, and nowhere else.
Daniel Fischer
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if we take a=0 then how we prove this function not differentiable at anot equal to 0 – asha Oct 22 '13 at 14:54
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Is your question how one proves that the function isn't differentiable at any other point than $a$? – Daniel Fischer Oct 22 '13 at 14:56
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yes , this is my question. – asha Oct 22 '13 at 14:57
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1At a point $y \neq a$, the function $h(x) = \dfrac{1}{x-a}$ is differentiable. So if $f$ were differentiable in $y$, then the function $h(x)\cdot f(x) = g(x)$ would also be differentiable in $y$. But $g$ is nowhere differentiable. – Daniel Fischer Oct 22 '13 at 15:05