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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.

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  • $\begingroup$ Hint: $g'(x) =-3\frac{x^2-5}{(5+x^2)^2}$ $\endgroup$ – gammatester Dec 9 '15 at 11:24
  • $\begingroup$ Is that the derivative of $g(x)$? Do you know a method which does not involve any calculus? $\endgroup$ – StopReadingThisUsername Dec 9 '15 at 11:27
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    $\begingroup$ @Arjun: Can you do calculus without doing calculus!? :) $\endgroup$ – H. R. Dec 9 '15 at 11:31
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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 $g$.

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  • $\begingroup$ I'm sorry, I didn't quite get that. Where did you get the $x = \frac{5}{y}$ in the first line and consequently the $g(x) = g( \frac{5}{x})$ in the second line? $\endgroup$ – StopReadingThisUsername Dec 9 '15 at 11:44
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    $\begingroup$ I solved the equation $g(x)=g(y)$. A function is injective if $g(x)=g(y) \implies x=y$. So, I solved $g(x)=g(y)$. The second line is just substitution. In the first line I found that if $x=\frac5y$ then $g(x)=g(y)$. So for $y=\frac5x$ I have that $g(x)=g(5/x)$. $\endgroup$ – Jimmy R. Dec 9 '15 at 11:48
  • $\begingroup$ I apologise if I'm missing something very obvious, but I don't understand how you deduced that $x = \frac{5}{y}$ as a solution to $g(x) = g(y)$... $\endgroup$ – StopReadingThisUsername Dec 9 '15 at 11:54
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    $\begingroup$ $g(x)=g(y) \iff \frac{3x}{5+x^2}=\frac{3y}{5+y^2}\iff yx^2-(5+y^2)x+5y=0 \iff \ldots$. Quadratic equation in $x$... $\endgroup$ – Jimmy R. Dec 9 '15 at 11:55
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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 $a = \sqrt{5}$.

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