So, I have a function $f(x) := \frac{3x}{7+x^2}$ and the question is that the domain is defined such as $x \geq a$, find the minimum value for $a$ for which there exists an inverse of the function.

On graphing this function, we get a S-Graph like this:

Like this

Via analysis, it's pretty evident that for $-2 \leq a \leq 2$. How do I find that algebraically?

  • $\begingroup$ are you sure about your $a$? the domain of $f$ is $[a, \infty)$. $\endgroup$
    – user251257
    Sep 8, 2015 at 14:00

1 Answer 1


First you have to calculate the derivative fo $f(x)$. This turns out to be $$f'(x)=\frac{21-3x^2}{(7+x^2)^2}$$

Now remember that you can calculate the inverse of a function as long as it is strictly monotonous, or expressed in terms of the derivative $f'(x) > 0$ or $f'(x)< 0$.

As you can see the denominator of the derivative is always positive. Hence only the numerator can change its sign. By calculating its zeros (place where the sign of the derivative changes) at $x_{1/2}=\pm \sqrt{7}$ you can conclude that the function is invertable for $x \in [\sqrt{7},\infty)$, $x \in [-\sqrt{7},\sqrt{7}]$ and $x \in (-\infty,-\sqrt{7}]$.

Using elementary math: Lets say we pick $x_1$ and $x_2$ with $x_1<x_2$. Then lets have a look at $f(x_2)-f(x_1)$. $$f(x_2)-f(x_1)=\frac{3x_2}{7+x_2^2}-\frac{3x_1}{7+x_1^2}=\frac{21x_2-21x_1+3x_2x_1^2-3x_1x_2^2}{(7+x_2^2)(7+x_1^2)}$$ $$=\frac{21(x_2-x_1)-3x_1x_2(x_2-x_1)}{(7+x_2^2)(7+x_1^2)}=\frac{(x_2-x_1)(21-3x_1x_2)}{(7+x_2^2)(7+x_1^2)}$$

As the denominator is positive and $x_2-x_1$ is also positive because of $x_1<x_2$, only $21-3x_1x_2$ can become zero. so lets try to solve this for $0$. $$21-3x_1x_2=0$$ $$x_1x_2=7$$ $$x_1=\frac{7}{x_2}$$

Lets plug this relation into the inequality $x_1<x_2$.


For $x_2$ positive we get $$7 < x_2^2$$

For $x_2$ negative we get $$7>x_2^2$$

Solving these inequalities will give you the same result as in the previous calculations.

  • $\begingroup$ Sadly, we haven't done calculus and we can't really use that method; any other method for pre-calculus? $\endgroup$
    – weirdpanda
    Sep 8, 2015 at 14:39
  • $\begingroup$ Added an pre-calculus calculation. $\endgroup$
    – MrYouMath
    Sep 8, 2015 at 15:02

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