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I recently learnt the proof that if a function is differentiable then it is continuous and that a function cannot be differentiable at a point discontinuity . I am confused however take the function $$ f(x)=\begin{cases} -2x & x<4, \\ 4 & x=9. \end{cases} $$ Isn't this function differentiable at $x=9$ as the left hand derivative exsists and the right hand derivative does not need to exsist as the function is not even defined for values of x greater than $9$ so what is the function differentiable and if not why?

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  • $\begingroup$ Did you mean $x<9$? $\endgroup$ Dec 30 '20 at 17:06
  • $\begingroup$ Sure your function is right? As written, f is not defined between 4 and 9 so it makes no sense to talk about it being differentiable or continuous at x = 9. $\endgroup$
    – Paul
    Dec 30 '20 at 17:09
  • $\begingroup$ But isn't that the purpose of piecewise functions $\endgroup$ Dec 30 '20 at 17:11
  • $\begingroup$ For calculating the left hand derivative, you would need the value of the function at the left of $9$. Remember that LHD is given by $$\lim_{h\to0}\frac{f(9)-\color{red}{f(9-h)}}h$$ $\endgroup$ Dec 30 '20 at 17:14
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    $\begingroup$ As written, this function is continuous. For $x<4$, choose $\delta<\frac{\varepsilon}{2}$. For $x=9$, choose $\delta<5$, not depending on $\varepsilon$. But it's not differentiable at $9$, since differentiability only makes sense at limit points of the domain, and $9$ is not a limit point of $(-\infty,4)\cup\{9\}$. $\endgroup$ Dec 30 '20 at 17:35
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This function is continuous but not differentiable at 9, this all comes down to the function's domain which is $(-\infty, 4)\cup \{9\}$. Here, 9 is an isolated point, and by the definition of continuity, every function is continuous at the isolated points of its domain. Moreover, as 9 is an isolated point, it is not an accumulation point, so again by definition you cannot calculate the limit $\lim_{h\rightarrow 0}\dfrac{f(9+h)-f(9)}{h}$. In fact, most books only define the derivative at a point when it is an interior point of the domain.

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    $\begingroup$ You may wish to prepare yourself for a moving target. I suspect that the OP intended that $f(4) = 9,$ not $f(9) = 4.$ $\endgroup$ Dec 30 '20 at 17:52
  • $\begingroup$ Oh, I see. In that case, f is not differentiable at 4, because $\lim_{h\rightarrow 0^-}\dfrac{f(4+h)-f(4)}{h}=\lim_{h\rightarrow 0^-}\dfrac{-8-2h-9}{h}=\lim_{h\rightarrow 0^-}\dfrac{-17}{h}-2=+\infty$. $\endgroup$ Dec 30 '20 at 18:06
  • $\begingroup$ right, but arguably premature. examining the wording of the OP's query: [1] on the one hand, his inclusion of an isolated point in the domain is bizarre, and the inferred typo is plausible [2] "Isn't this function differentiable at $x=9$ as the left hand derivative exists" : he specifically referred to $x=9$ : is this another typo? [3] he said : "as the left hand derivative exists" : exists where, at $x=9$?? You may wish to hold off firing torpedoes until the target stops moving. $\endgroup$ Dec 30 '20 at 18:13
  • $\begingroup$ @bruno aceves can't a function be differentiable at end points however isn't saying that the derivative is only defined for interior points incorrect as we can use the lateral derivative at the end points? $\endgroup$ Jan 1 at 17:09
  • $\begingroup$ @Thehomeschooler there are some texts that define the derivative at end points of as you say and some that only define it in interior points. The problem here is that you cannot even calculate the limit that defines the derivative at 9, because 9 must be an accumulation point to do that. $\endgroup$ Jan 1 at 20:03

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