2
$\begingroup$

I am trying to show that this function $s:\mathbb R→\mathbb R$ defined by $s(x)=e^x - e^{-x}$ is one-to-one.

My approach is using the natural $\log$ ($\ln$) to cancel out the $e$'s: $$e^x - e^{-x} = e^y - e^{-y}\\ \ln(e^x - e^{-x}) = \ln(e^y - e^{-y})\\ x + x = y + y\\ 2x = 2y\\ x = y$$ I just wanted to know:

  • if my algebra is correct and you can do the above operations
  • if it is not correct, would the next best way to prove the one-to-one property by just plugging in zero for the original functions?

Thanks.

$\endgroup$
8
  • 5
    $\begingroup$ The algebra is not right. The ln of a sum is not the sum of the ln's. $\endgroup$ Mar 14, 2013 at 21:18
  • $\begingroup$ What sort of tools do you have? I assume you're not supposed to use calculus? $\endgroup$ Mar 14, 2013 at 21:22
  • $\begingroup$ Yes I think it should be done using basic algebra, but if there is a simple way to do prove x = y using calculus I'd be enlightened to see it as well. $\endgroup$
    – Goose
    Mar 14, 2013 at 21:23
  • 1
    $\begingroup$ A simple calculus-proof would be: Let $f(x) = e^x - e^{-x}$. Then $f'(x) > 0$ for all $x$, so $f$ is strictly increasing and thus $f(x) = f(y)$ implies $x = y$. $\endgroup$
    – TMM
    Mar 14, 2013 at 21:32
  • $\begingroup$ @TMM thank you that is pretty simple and makes sense, I'd still like an algebraic one if possible. Gerry Myerson, thanks I just edited my question. $\endgroup$
    – Goose
    Mar 14, 2013 at 21:39

3 Answers 3

7
$\begingroup$

$$\begin{align} e^x - e^{-x} &= e^y - e^{-y} \\ e^x + e^{-y} &= e^y + e^{-x} \\ e^x(1 + e^{-x-y}) &= e^y(1 + e^{-x-y})\end{align}$$ Since $e^{-x-y} \geq 0$ for all $x,y$ we have $1 + e^{-x-y} \neq 0$, so we can cancel the terms on both sides to get $$e^x = e^y$$ Since $e^x$ is injective, the result follows.

$\endgroup$
6
$\begingroup$

Perhaps it is enough to observe that since $e^x$ is strictly increasing $x>y \Rightarrow e^x > e^y$

so

$e^{-x} < e^{-y}$

$-e^{-x} > -e^{-y}$

$e^x -e^{-x} > e^y -e^{-y}$

$\endgroup$
2
$\begingroup$

If $e^x-e^{-x}=w$, then $e^{2x}-we^x-1=0$, a quadratic equation in $e^x$. By the quadratic formula, $e^x=(w\pm\sqrt{w^2+4})/2$. We reject the solution with the minus sign, since it makes $e^x$ negative, which is impossible. So $e^x=(w+\sqrt{w^2+4})/2$, and $x=\log((w+\sqrt{w^2+4})/2)$, a unique solution. So $f(x)=e^x-e^{-x}$ is one to one.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .