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I learned that all differentiable functions are continuous. Why does the following equation not violate this rule:

$$f(x)=\begin{cases}x^2+3 \quad &\text{when } x>1 \\ x^2 \quad &\text{when }x\le 1\end{cases}$$

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  • $\begingroup$ The derivative at $x = 1$ does not exist as $\lim_{x\to 1} \frac{f(x) - f(1)}{x-1} = \lim_{x\to 1} \frac{f(x) - 1}{x-1}$ does not exist. In particular, the limit from above. $\endgroup$ – Simon S Dec 7 '14 at 4:12
  • $\begingroup$ That function is neither continuous nor differentiable at $x = 1$. $\endgroup$ – Daniel Hast Dec 7 '14 at 4:12
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    $\begingroup$ I am confused the derivative of x^2+3 = 2x and the derivative of x^2 = 2x where is the error $\endgroup$ – Matthew Baranoff Dec 7 '14 at 4:18
  • $\begingroup$ See math notation guide. $\endgroup$ – user147263 Dec 7 '14 at 4:19
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    $\begingroup$ but the two side both equal 2x do they not $\endgroup$ – Matthew Baranoff Dec 7 '14 at 4:34
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$f(x)$ is not differentiable at $x = 1$. The limit $$\lim_{x\rightarrow 1+}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow 1+}\frac{x^2+2}{x-1}$$ does not exist.

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  • $\begingroup$ Great use of algebra to demonstrate the fact that the limit of the difference quotient does not exist. What might add to one's conceptual understanding is a geometric explanation: the slope of the line between $\bigl(1, f(1)\bigr)$ and $\bigl(1+h, f(1+h)\bigr)$ increases without bound (the line gets steeper) as $h\to 0$. See this graph for a demo. $\endgroup$ – chharvey Dec 7 '14 at 21:42
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This function is not differentiable at $x=1$, nor is it continuous there. That's why.

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  • $\begingroup$ It would be nice if this answer demonstrated why the function is not differentiable at 1. Also, the fact that it is discontinuous there is unnecessary. The OP knows it is discontinuous, which is why they asked the question in the first place: it was an attempt at a counterexample. $\endgroup$ – chharvey Dec 7 '14 at 21:33

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