1

Range of $f(x) = \frac{1+e^x}{1-e^x}$

To understand more about the range, I calculated the domain of $f(x)$ which is all real numbers except $x=0$

From the domain, I understand that there is no values of $f(x)$ when $x=0$

I had a hint from someone else that $0< e^x <1$ or $1<e^x$

I am not sure what this hint suppose to mean in finding the range.

user307640
  • 2,632
  • 2
  • 17
  • 27
  • 2
    Have you tried sketching the function? – Mark Bennet Sep 09 '22 at 16:40
  • Just study the variations and the limits.. – Lelouch Sep 09 '22 at 16:45
  • If you're familiar with hyperbolic functions, you could write$$f(x) = \frac{1+e^x}{1-e^x} = \frac{e^{-x/2}+e^{x/2}}{e^{-x/2}-e^{x/2}} = -\coth\left(\frac x2\right)$$ – user170231 Sep 09 '22 at 16:46
  • 1
    Since $e^x$ is continuous, the function is going to be continuous except for where it is undefined (when $1-e^x=0$), so the range will be a collection of intervals. So ask yourself, what happens when $e^x$ gets closer to $0$? When it gets close to $1$ but is still smaller? When it gets closer to $1$ but is bigger? When it goes off to infinity? – Aaron Sep 09 '22 at 16:50

2 Answers2

3

We have $$f(x)=\frac{1+e^x}{1-e^x}$$ with domain $$\mathbb{D}_{f(x)}=(-\infty, 0)\cup(0,\infty)$$ If you take the derivative to find extrema, $$f'(x)=\frac{2e^x}{\left(1-e^x\right)^2}=0$$ this has no solutions. There are not extrema so $f$ should smoothly transition from an upper bound to lower bound along the endpoints of both parts of the domain.

Taking limits at the origin, we see that $$\lim_{x\to0^+}f(x)=-\infty$$ $$\lim_{x\to0^-}f(x)=\infty$$ Now, take the limits out to infinity, we see that $$\lim_{x\to\infty}f(x)=-1$$ $$\lim_{x\to-\infty}f(x)=1$$

So the range is just $$\mathbb{R}_{f(x)}=(-\infty, -1)\cup(1, \infty)$$

Max0815
  • 3,505
  • 1
    +1. My edit was to change $\in$ to $=$ in the last line. We can save a little work by observing that $f(-x)=-f(x)$ when $x\ne 0.$ – DanielWainfleet Sep 09 '22 at 17:26
  • @DanielWainfleet Aight that seems like better notation. i should probably change the domain one too then lol – Max0815 Sep 09 '22 at 17:27
0

I know this is not the correct thing to do; but I don't have the reputation to comment so I'll post it as an answer.

You can take the inverse of the function and the domain of the inverse will be the range of the original function.

$$f(x)=\frac{1+e^x}{1-e^x}$$ $$f^{-1}(x)=\frac{1+e^y}{1-e^y}$$ $$x=\frac{1+e^y}{1-e^y}$$ $$x(1-e^y)={1+e^y}$$ $$x-xe^y={1+e^y}$$ $$-xe^y-e^y={1-x}$$ $$e^y(-x-1)={1-x}$$ $$e^y=\frac{1-x}{-x-1}$$ $$y=\ln\left(\frac{1-x}{-x-1}\right)$$

Hope this helps.

Theodore Veteto
  • 107