Range of function $y=(e^x-e^{-x})/(e^x+e^{-x})$

Question

What will be the range of function $y=(e^x-e^{-x})/(e^x+e^{-x})$ ? How should I approach this category of problems with exponential functions? Please Help.

Just to verify, you are talking about the function otherwise described as $y=\tanh x$, yes? — abiessu, Apr 13 '15 at 18:39
You are right.But actually I don't really know much about hyperbolic functions,so for the time being lets not use tanh x. :-P — , Apr 13 '15 at 18:41
Solve that equation for $x$ and you'll see what are the valid values of $y$. — Simon S, Apr 13 '15 at 18:42
It's very useful. Building on the algebraic step Sammy just gave
$$e^{2x} = \frac{1+y}{1-y}$$

which necessarily means that $\frac{1+y}{1-y} > 0$. We have immediately that $y < 1$ and $y > -1$. There is a value of $x$ that attains every value of $y$ in that open interval, hence the range is $(-1, + 1)$. — Simon S, Apr 13 '15 at 18:48
@SimonS: you're right, I was under the mistaken impression that it was difficult to do such an inversion. — abiessu, Apr 13 '15 at 18:49
@SanchayanDutta, look again at what I've written. Same thing. — Simon S, Apr 13 '15 at 18:50

score 5 · Accepted Answer · answered Apr 13 '15 at 18:53

One way to analyze the possible values the given function can take on is to transform it into the "simplest possible" form. Some arithmetic we might apply could look like this:

$$y=\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ =\frac {e^{2x}-1}{e^{2x}+1}\\ =1-\frac 2{e^{2x}+1}$$

We could go further with this directly and fully invert the function, but we can immediately make some conclusions from where we are now:

$e^{2x}+1\in(1,\infty)$
$\frac 2{e^{2x}+1}\in(0,2)$

This leads us to $y\in(-1,1)$.

score 1 · Answer 2 · edited Sep 23 '17 at 03:18

1

You have $y=f(x)$. Solve for $x=g(y)$. Now find the domain for $g(y)$. That would be your "range".

edited Sep 23 '17 at 03:18

Parcly Taxel

103,344

answered Apr 13 '15 at 18:43

Nicolas Bourbaki

5,064

Range of function $y=(e^x-e^{-x})/(e^x+e^{-x})$

2 Answers2