Inverse function for $f(x) = \frac{\sqrt{x}}{\sqrt{2x+2}}$

Question

Let $f: [0, \infty) \to \mathbb{R}$ and $ f(x) = \frac{\sqrt{x}}{\sqrt{2x+2}}$ for every $x \in[0, \infty).$

Show that $f$ has an inverse function $f^{-1}:f([0, \infty)) \to [0, \infty).$ Determine the domain of $f^{-1}$. Compute $(f^{-1})'(\frac12).$

To show the existence of $f^{-1}$ my idea was to show that it's bijective, however computing $f'$ results in $f'(x) = \frac{1}{\sqrt{x}(2x+2)^{\frac32}}$ which is not greater than $0$ when $x=0$, hence $f$ is not strictly increasing and therefore not injective. Also $\lim_{x\to0} f(x)=0$, but $\lim_{x\to\infty} f(x)= \sqrt{\frac12}$ so $f$ is not even surjective(?). What am i not seeing here?

Answer 1

Your argument has flaws. For example $x^{3}$ is strictly increasing in $[0, \infty)$ even though the derivative at $0$ is not $>0$. This function is also injective.

Hints: Solving the equation $f(x)=y$ is a better approach than using monotonicity. The range of the function is $[0,\frac 1 {\sqrt 2})$ and the equation $f(x)=y$ has the unique solution $x=\frac {y^{2}} {1-2y^{2}}$ for $0 \leq y <\frac 1 {\sqrt 2}$. Since $\frac {y^{2}} {1-2y^{2}}$ is continuous on $[0,\frac 1 {\sqrt 2})$ the result follows.

Answer 2

$f(x)=\frac{1}{\sqrt{2}}\sqrt{\frac{x}{x+1}}$.

$\frac{x}{x+1}$ is strictly increasing, thus $\sqrt{\frac{x}{x+1}}$ and $f(x)$ are too... therefore they have inverse.

btw $f'(x)= \frac{\frac{1}{2}\sqrt{\frac{2x+2}{x}}-\sqrt{\frac{x}{2x+2}}}{2x+2}$ which is always greater than $0$ in the domain