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I have $F:\mathbb{R}^{*}\rightarrow \mathbb{R}$ such that $F'(x)=\frac{1}{x}$, for all $x\in \mathbb{R}^{*}, F(-1)=1$ and $F(1)=0$.I need to find the value of $F(e)+F(-e)$ (3 is the right answer)

If $F'(x)=1/x$ then $f(x)=1/x$ so $F(x)=ln|x|$ this mean that the result is 2 but it's wrong.How to start?

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    $\begingroup$ First step: Your conclusion that $F(x) = \ln |x|$ is mistaken; all you know is that $F(x) -= \ln |x| + C$ for some constant $C$...and more important, you only know this on each domain where $F$ is continuous, i.e., the $C$ for the negative reals might be different from the one for the positive reals. $\endgroup$ – John Hughes Apr 30 '19 at 19:43
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The problem with what you have done is the fact the $\mathbb{R}^*$ is not an interval. You have to split it into $(0,\infty)$ and $(-\infty,0)$. In both interval $F$ has primitives, those two primitives are not necessary the same.


Here’s is something you can use

If $x>0$ then $F(x)=\ln x+c_1$ since $F(1)=0$ then $c_1=0$.

If $x<0$, then $-F’(x)=\frac1{-x}$ which means $F(x)=\ln(-x)+c_2$. Now: $F(-1)=c_2=1$ thus $F(-x)=1+\ln(-x)$

$F(e)=\ln e=1$ and $F(-e)=1+\ln(e)=2$ therefore $F(e)+F(-e)=1+2=3$

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  • $\begingroup$ Thank you for your help:) $\endgroup$ – DaniVaja May 1 '19 at 15:53

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