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Let $p(x)$ and $q(x)$ be two differentiable function on $\mathbb{R}$. Now how to check the differentiability of the following function at $x=a$

$$ f(x) = \begin{cases} p(x), & \mbox{if } x\mbox{ $\geq a$} \\ q(x), & \mbox{if } x\mbox{ $ <a$} \end{cases} $$?. I usually check it as , if $\lim_{x\to a^{+}} p^{'}(x)$=$\lim_{x\to a^{-}} q^{'}(x)$ then $f$ is differentiable at $a$. Am i right in this calculation.? Please suggest me is it right method to check differentiability of the function $f$ at $x=a.$?

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  • $\begingroup$ Please some one suggest me... $\endgroup$ – Parvesh Kumar May 15 '15 at 5:04
  • $\begingroup$ No. You must check that $\lim_{x \to a^+} \frac{p(x)-p(a)}{x-a} = \lim_{x -\to a^-} \frac{q(x)-p(a)}{x-a}.$ $\endgroup$ – matt biesecker May 15 '15 at 5:12
  • $\begingroup$ You need to check differentiability of $f$ not continuity of $p',q'$. $\endgroup$ – copper.hat May 15 '15 at 5:13
  • $\begingroup$ @mattbiesecker if q(x) is also defined at x=a then the method of parvesh kumar will work?? $\endgroup$ – neelkanth May 15 '15 at 6:17
  • $\begingroup$ @Yogesh. If $q(a)$ is defined, then it must equal $p(a),$ otherwise $f$ will not even be a function. $\endgroup$ – matt biesecker May 15 '15 at 6:23
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At a minimum you need continuity. What condition is necessary at $x=a$ for continuity?

You need $\lim_{x \downarrow a} {f(x)-f(a) \over x-a} = \lim_{x \uparrow a} {f(x)-f(a) \over x-a}$.

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To specifically answer why the OP's method is insufficient, consider the following scenarios.

1) A case where the derivative exists, but the left/hand hand limits do not exist is $$f(x) = \left\{ \begin{array}{cl} x^2 \sin(1/x) \ \ & x\neq 0 \\ 0 & x=0 \end{array} \right.\ \ .$$ Then both of $\lim_{x\to 0^+} f'(x)$ and $\lim_{x\to 0^-} f'(x)$ do not exist. However, $f'(0)=0$ via the definition of differentiability.

2) A case where the left and right hand limits of $f'$ agree, but $f'(0)$ does not exist is rather trivial:
$$ f'(x) = \left\{ \begin{array}{cl} 1 \ \ & x\geq 0 \\ 0 & x<0 \end{array} \right.\ \ . $$ Then $\lim_{x\to 0^+} f'(x) = \lim_{x\to 0^-} f'(x)=0,$ but $f'(0)$ does not exist.

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  • $\begingroup$ nicely explaned...thanks a lot... $\endgroup$ – neelkanth May 15 '15 at 6:57

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