Find injectivity/ prove one to one algebraically of $ \frac{|x|x}{\sqrt{x^4-81}}$

Question

I'm trying to find the range of this function: $$ \frac{|x|x}{\sqrt{x^4-81}}. $$ However to find the range I have to first prove one to one, then find domain of inverse. I can't prove injectivity without restricting domain algebraically. How can I do this?

Answer 1

Let's denote $f$ this function. Observe $f$ is odd and $f(x)$ (when defined) has the sign of $x$. Hence it is enough to prove injectivity and find the range when $x>0$.

Now $f(x)$ is defined if and only $x^4>81\iff x^2>9\iff \vert x\rvert>3$. Thus the domain of $f$ is $(-\infty,-3)\cup(3,+\infty)$.

Finally let's determine when $f$ is monotonic. Let $x,y>3$ and set $X=x^2, Y=y^2$; we have: \begin{align*} f(x)<f(y)&\iff \frac X{\sqrt{X^2-81}} <\frac Y{\sqrt{Y^2-81}}\iff \frac{X^2}{X^2-81} <\frac{Y^2}{Y^2-81}\\ &\iff X^2(Y^2-81)< Y^2(X^2-81)\iff Y^2<X^2\\ &\iff y^4<x^4\iff y<x. \end{align*} Hence $f$ is decreasing on $(3,+\infty)$.

Note: It is slighly faster to use calculus, since if $x>0$, $$f'(x)=\frac{2x\sqrt{x^4-81}-x^2\dfrac{4x^3}{2\sqrt{x^4-81}}}{x^4-81}=\frac{-162x}{(x^4-81)^{3/2}}.$$