Inverses of Rational Functions

Question

Consider the function $g$, where $$g(x)=\frac{3x}{5+x^2}$$

(a) Given that the domain of $g$ is $x\ge a$, find the least value of $a$ such that $g$ has an inverse function.

I know that $g(x)$ must be a one-to-one function, thus pass the horizontal line test, for an inverse to exist. This is a non-calculator question; I tried to sketch the graph by inputting random values to get the general trend, to determine particularly the local maximum, after which the function would decline and could be a one-to-one function, hence giving us the value of $a$. However, this method proved to be very time consuming - I was wondering if there was a faster technique to do so, if possible, without calculus.

Any help will be greatly appreciated, thanks in advance.

Answer 1

Hint: $$g(x)=g(y) \iff x=y \quad \text{ or } \quad x=\frac5y$$ Indeed you can verify that $g(x)=g\left(\frac{5}{x}\right)$. So, you must not allow that $x$ and $5/x$ are in the domain of $g$.

Answer 2

You can sketch the graph by analysing the extrema of $g(x)$, dispense with the need to get the graph point by point.

Notice that

\begin{align} \frac{1}{g(x)} &= \frac{5+x^2}{3x} = \frac{1}{3} \left( \frac{5}{x} + x \right) \\ &\ge \frac{2\sqrt{5}}{3} \end{align}

Equality holds if and only if $x = \pm\sqrt{5}$.

It is easy to sketch the graph of $\frac{1}{g(x)}$, and to see that the graph of $\frac{1}{g(x)}$ is a hyperbola, between $x = 0$ and $y = \frac{1}{3}x$, with vertices at $x = \pm \sqrt{5}$.

Then it's easy to sketch the graph of $g(x)$, which is just the reciprocal of the above.

Then you can see that $g(x)$ get to the maximum at $x = \sqrt{5}$, and when $x \ge \sqrt{5}$, the curve declines continuously. So you can get $a = \sqrt{5}$.