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I have a question regarding the theorem 'If $f$ is differentiable at a point $x$, then $f$ must also be continuous at $x$'.

Take, for example, following function: $$f(x) = \begin{cases} 3x^2 &\mbox{if } x \leq 2\\ x^3+1 &\mbox{if }x >2\\ \end{cases}$$

If you look at the differentiability for $x=2$, we see that both the left and right derivative is equal to $12$. So $f'(2)=12$, so I conclude that $f$ is differentiable for $x=2$. However, the function is clearly not continuous for $x = 2$, which contradicts the theorem. I already looked to the proof but I don't find any need for extra conditions in it (for example, if a function $f$ is differentiable in $(a,b)$, then $\forall c \in (a,b)$ $f$ is also continuous in $c$).

What is wrong in my interpretation of the theorem or my interpretation of the definition of being differentiable at a point?

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  • $\begingroup$ What is $(f(2 + \epsilon) - f(2))/\epsilon$ for small positive $\epsilon$? $\endgroup$ – andars Jun 11 '17 at 3:05
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    $\begingroup$ $\lim_{x \to 2^+} f'(x)$ and $\lim_{h \to 0^+} \frac{f(2+h)-f(2)}{h}$ are not the same in this case. $\endgroup$ – Ian Jun 11 '17 at 3:07
  • $\begingroup$ You are right, sorry this was a typo. I changed my question with a correct example. $\endgroup$ – Filip M Jun 11 '17 at 3:07
  • $\begingroup$ Same problem, the right derivative does not exist at $2$. $\endgroup$ – Ian Jun 11 '17 at 3:08
  • $\begingroup$ As Zachary Selk explains below, you are making a mistake. You may have seen other cases similar to this where the function does turn out to be differentiable. For example, let $f(x) = x^2$ when $x \geq 0$ and $f(x) = -x^2$ when $x < 0$. The right-hand derivative $f_+'(0) = 0$, because the function $f(x)$ agrees with $x^2$ for all $x \geq 0$. Likewise, the function $f(x)$ agrees with $-x^2$ for all $x \leq 0$ (not just $x < 0$), so $f_-'(0) = 0$. This is where the analogy with your example breaks down. $f(x) = x^3$ is correct for all $x > 2$, but not for all $x \geq 2$. $\endgroup$ – user49640 Jun 11 '17 at 3:17
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The left and right hand derivatives are NOT $12$. Well the left hand derivative is, but the right hand doesn't exist. Note that the right hand derivative is:

$$\lim_{h\to 0^+} \frac{f(2+h)-f(2)}{h}$$

Note this is NOT the same thing as:

$$\lim_{x\to 2^+}f'(x)$$

These are two different things.

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  • $\begingroup$ You are right, sorry this was a typo. I changed my question with a correct example. $\endgroup$ – Filip M Jun 11 '17 at 3:07
  • $\begingroup$ @FilipM My answer still stands $\endgroup$ – user223391 Jun 11 '17 at 3:08
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    $\begingroup$ @FilipM For an easier example, what are the left and right hand derivatives of the function $f(x)=\begin{cases} 1 & \text{ if } x>0\\ 0 & \text{ if } x \le 0\end{cases}$ $\endgroup$ – user223391 Jun 11 '17 at 3:17
  • $\begingroup$ Thank you for your answer. I understand that there is something wrong with he right hand derivative and that it not just calculating (x^3)' and evaluating it in x = 2. But why exactly does the expression $\lim_{h\to0^+}\frac{f(2+h)−f(2)}{h}$ not exist? $\endgroup$ – Filip M Jun 11 '17 at 3:38
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    $\begingroup$ @FilipM Try calculating it. The top goes to $2^3-3\times 2^2=-4$ and the bottom goes to $0$, in other words the limit doesn't exist $\endgroup$ – user223391 Jun 11 '17 at 3:42

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