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?

  • 1
    $\begingroup$ 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

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 $F(e)+F(-e)=1+2=3$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.