Given a function find its inverse and solve the inequality $2f(x^2-1)<1$

Question

Given $$f(x)={e^x\over e^x+1}, x\in \mathbb R$$

•$D_{f^-1}=(0,1)$

•$f$ continuous at $\mathbb R$

•$f$ strictly increasing at $\mathbb R$

I) Find the inverse of $f.$

II) Solve the inequality: $$2f(x^2-1)<1.$$

Personal work:

I) $$f(x)=y\iff {e^x\over e^x+1}=y\iff y(e^x+1)=e^x$$ now do I solve for $e^x$ and then apply $\ln $ to both sides of the equation?

ΙΙ) $$2f(x^2-1)<1\iff f(x^2-1)<{1\over 2}=f(0)$$ (1)

$$\forall x\in D_f: x^2>0\iff x^2-1>-1$$

$$\forall x \in D_f:0\not > 0$$

$$(x^2-1)\in\mathbb R$$ where $f$ is strictly increasing. Therefore, (1)$\iff x^2-1<0\iff x^2<1\iff \boxed{-1<x<1}.$

Answer 1

Hint: write $$e^x+1=\frac{e^x}{y}$$ and then $$1=e^x\left(\frac{1-y}{y}\right)$$ so $$e^x=\frac{y}{1-y}$$ Can you finish?

Answer 2

Note that

$${e^x\over e^x+1}=y\iff \frac{e^x+1}{e^x}=1+\frac1{e^x}=\frac1y$$

Answer 3

For $y\in (0,1),$ we look for $x\in\Bbb R $ such that

$$e^x (1-y)=y \iff e^x=y/(1-y) $$

$$\iff x=\ln \Bigl (\frac {y}{1-y}\Bigr)=f^{-1}(y)$$

for the second

$$2f (x^2-1)<1\iff f (x^2-1)<f (0) $$ $$\iff x^2-1 <0\iff -1 <x <1$$