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$f(x) \begin{cases} \frac{x^2+4x}{1-2x} , x<0\\ \int^x_0 e^{cos(t)} , x\geq 0\\ \end {cases} $ is the function differentiable in all $\Bbb R$


start of with with the case $x<0$ the function is differentiable and continuous for all $x<0$ as an elementary function and the same for $x\geq 0$ case

the only problem seems to be the point where $x=0$

so I decided to check the derivative by definition at the given point

$\lim \limits_{x\to 0} \frac{f(x)-f(0)}{x-0}$ and according to the fundamental theorem of calculus for the case $x \geq 0$ we have $F'(x)=(\int^x_0 e^{cos(t)})'=e^{cosx}$ so $\lim \limits_{x\to 0} \frac{f(x)=e^{cosx}-f(0)=0}{1-0}$ (according to lhopital and the the fundamental theorem) so the derivative is $e$

so I get that it is differentiable , is that correct?

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1 Answer 1

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No, it is not differentiable at $0$, since$$\lim_{x\to0^-}\frac{f(x)-0}x=4\quad\text{and}\quad\lim_{x\to0^+}\frac{f(x)-f(0)}x=e^{\cos(0)}=e$$and $4\ne e$.

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  • $\begingroup$ but aren't the side limits of $f(x)$ both 0? in the first case we get $\frac{0}{1}$ and the second is just an integral from $0$ to $0$ $\endgroup$
    – Adamrk
    Commented Jan 30, 2023 at 16:20
  • $\begingroup$ Yes, $F(0)=0$; that was a typo. But$$\lim_{x\to0^-}\frac{x^2+4}{1-2x}=\frac{0^2+4}{1-2\times0}=\frac41=4.$$ $\endgroup$ Commented Jan 30, 2023 at 16:22
  • $\begingroup$ I edited it , sorry it was a typo $\endgroup$
    – Adamrk
    Commented Jan 30, 2023 at 16:23
  • $\begingroup$ I have adapted my answer to the new version of your question. The new version of the function is indeed continuous at $0$. $\endgroup$ Commented Jan 30, 2023 at 16:27
  • $\begingroup$ @so is it still not differentiable? $\endgroup$
    – Adamrk
    Commented Jan 30, 2023 at 16:28

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