Prove that $f(x)=\log\sqrt{\frac{1+x}{1-x}}$ is surjective from $(-1,1)$ to $\mathbb{R}$. I have to prove that the function $\;f:(-1,1)\to \mathbb{R}\;$ defined by $f(x)=\log\sqrt{\frac{1+x}{1-x}}\;$ is bijective.
I have already proved that it is injective:
$$f(x)=f(y)$$
$$\log\sqrt{\frac{1+x}{1-x}}=\log\sqrt{\frac{1+y}{1-y}}$$
$$\log\sqrt{\frac{(1+x)(1-y)}{(1-x)(1+y)}}=0$$
$$\frac{1+x-y-xy}{1-x+y-xy}=1$$
$$x=y$$
But now, how can I prove that the function is surjective?
 A: We have that $f(x)$ is defined in $(-1,1)$ and
$$f(x)=\log\sqrt{\frac{1+x}{1-x}}\implies f'(x)=\frac1{1-x^2}>0$$
then $f(x)$ is injective, moreover
$$\lim_{x\to 1^-} f(x)=\infty \quad \lim_{x\to -1^+} f(x)=-\infty$$
and since $f(x)$ is continuous by IVT it is surjective.
A: Let $y\in\mathbb{R}$. Then the equation
$$
y=\log\sqrt{\frac{1+x}{1-x}}
$$
becomes
$$
\frac{1+x}{1-x}=e^{2y}
$$
that solves as
$$
x=\frac{e^{2y}-1}{e^{2y}+1}
$$
This can be rewritten as $x=\tanh y$, but is not relevant. Note that
$$
-1<\frac{e^{2y}-1}{e^{2y}+1}<1
$$
if and only if
$$
-e^{2y}-1<e^{2y}-1<e^{2y}+1
$$
and both inequalities are obviously true for every $y$.
This can be simplified if you know about hyperbolic function: you need to prove that $-1<\tanh y<1$, that is, $\tanh^2y<1$ or $\sinh^2y<\cosh^2y$, which is clear because $\sinh^2y=\cosh^2y-1$.
The proof of injectivity can be shortened by noticing that both the logarithm and the square root are injective, so you just have to prove that
$$
\frac{1+x}{1-x}=\frac{1+y}{1-y}
$$
implies $x=y$. The given equality becomes
$$
1+x-y-xy=1+y-x-xy
$$
that's exactly $x=y$.
A: $f:(-1,1) \to \mathbb R^+, f(x) = \frac {1+x}{1-x}$ is continuous, monotonically increasing, and a bijection.
$g: \mathbb R^+\to \mathbb R^+, g(x) = \sqrt{x}$ is continuous, monotonically increasing, and a bijection.
$h: \mathbb R^+\to \mathbb R, h(x) = \ln x$ is continuous, monotonically increasing, and a bijection.
The composition of bijections gives a bijection.
You might also note that:
$\sqrt \frac{1+x}{1-x}$
substituting $x = \cos \theta$ gives
$\sqrt \frac {1+\cos \theta}{1-\cos\theta} = \cot \frac \theta2$
A: We can show that $f$ is surjective using a proof by contradiction and a few basic facts about $+, -, \exp, \log, \sqrt{\cdots} \dots $ .
$\renewcommand{isa}{\mathop{:}} \renewcommand{opdot}{\mathop{.}}$Our hypothesis that $f(x) = \log\sqrt{\frac{1+x}{1-x}}$ where $\text{-1} < x < 1$ can be written using quantifiers as:
$$ \forall y \isa \mathbb{R} \opdot \exists x \isa\, (\text{-1}, 1) \opdot \log \sqrt{\frac{1+x}{1-x}} = y \tag H $$
Let's take our hypothesis and negate it ... and then show that the negation is inconsistent.
$$ \exists y \isa \mathbb{R} \opdot \forall x \isa\, (\text{-1}, 1) \opdot \log \sqrt{\frac{1+x}{1-x}} = y \tag{NG1} $$
We can assume that we have a counterexample, let's name it $c$ . We only get one counterexample though, so after we assume (NG2), (NG1) can't be used anymore.
$$ \forall x \isa\, (\text{-1}, 1) \opdot \log \sqrt{\frac{1+x}{1-x}} \ne c \tag{NG2} $$
For the sake of conciseness, let's leave off $x \isa\, (\text{-1}, 1)$ on our intermediate steps.
$$ \log \sqrt{\frac{1+x}{1-x}} \ne c \tag{1} $$
We can apply an injective function $g$ to both sides of the inequality and get back another true inequality $a \ne b \iff g(a) \ne g(b)$. Let's start with $exp$ .
$$ \exp \log \sqrt{\frac{1+x}{1-x}} \ne \exp{c} \tag{2} $$
$exp$ is the inverse of $\log$
$$ \sqrt{\frac{1+x}{1-x}} \ne \exp{c} \tag{3} $$
The function that sends $x$ to its square is not injective on the reals. However, $\exp(c)$ is positive and $\sqrt{\cdots}$ is real and hence non-negative. Therefore in this particular case, squaring will preserve $(\ne)$ .
$$ \frac{1+x}{1-x} \ne \exp{2c} \tag{4} $$
multiplying by $1-x$ is injective regardless unless $x=1$, but $x$ is defined to be in $(\text{-1}, 1)$ .
$$ 1+x \ne (1-x)\exp{2c} \tag{5} $$
$$ 1+x \ne (\exp{2c}) - x\exp{2c} \tag{6} $$
Adding a constant $x\exp{2c}$ and $-1$ is injective.
$$ x + x\exp{2c} \ne (\exp{2c}) - 1 \tag{7} $$
Division by a positive constant $(\exp{2c})+ 1$ is injective
$$ x \ne \frac{(\exp{2c})-1}{(\exp{2c})+1} \tag{8} $$
$$ \bot $$
In order to produce a contradiction, we pick $x = \frac{(\exp{2c})-1}{(\exp{2c})+1}$ .
Our choice of $x$ lies in the open interval $(\text{-1}, 1)$, but we need to prove it.
In order to show $\text{-1} < x < 1$, we'll show that $x \le -1$ and $x \ge 1$ are absurd.
$$ \frac{(\exp{2c})-1}{(\exp{2c})+1} \le \text{-1} \tag{NG3} $$
multiplication by a positive constant is monotonic, preserve $(\le)$.
$$ (\exp{2c}) -1 \le -1 - \exp{2c} \tag{9} $$
addition by a constant is monotonic.
$$ (\exp{2c}) + \exp{2c} \le 0 \tag{10} $$
simplify and divide by two.
$$ \exp{2c} \le 0 \tag{11} $$
$$ \bot $$
Contradiction, because $\exp{2c}$ is positive.
Now to the other boundary case.
$$ \frac{(\exp{2c}) - 1}{(\exp{2c})+1} \ge 1 \tag{NG4} $$
$$ (\exp{2c}) - 1 \ge (\exp{2c}) + 1 \tag{12} $$
$$ 0 \ge 2 \tag{13} $$
$$ \bot $$
Contradiction.
Therefore, if there were any $c \in \mathbb{R}$ that weren't hit by $f$, then $\frac{(\exp{2c}) - 1}{(\exp{2c}) + 1}$, which is in $(\text{-1}, 1)$, wouldn't be in the domain of definition of $f$ . However, $f$ is surjective and therefore total.
