# A way of finding the range of an injective function

Let $f: [-1,3[\to \mathbb{R}$ be a function define by: $$f(x)=\sqrt{\frac{x+1}{3-x}}$$

My goal was to find the range of the function without using the inverse, $f(y)^{-1}$. This method, only works if the function is injective. If the function is not injective, it can have a maximum/minimum values above/below the endpoints of the domain. So lets look if $f(x)$ is an one-to-one function.

One way is to assume that $f(x)=f(y)$, for some $x,y \in [-1,3[$ and then prove that $x=y$. $$\sqrt{\frac{x+1}{3-x}}=\sqrt{\frac{y+1}{3-y}}\Leftrightarrow$$

$$\frac{x+1}{3-x}=\frac{y+1}{3-y} \Leftrightarrow$$

$$(x+1)(3-y)=(y+1)(3-x)\Leftrightarrow$$

$$3x-xy+3-y=3y-xy+3-x\Leftrightarrow$$

$$3x-y=3y-x\Leftrightarrow 3x+x=3y+y \Leftrightarrow$$

$$(3+1)x=(3+1)y \Leftrightarrow x=y _{\blacksquare}$$

Because, $x=y$ when $f(x)=f(y)$, then $f$ is injective.

So, now lets find the values of $f$ at the endpoints of the domain.

$$\lim_{x \to -1}\sqrt{\frac{x+1}{3-x}}=0$$

and,

$$\lim_{x \to 3^-}\sqrt{\frac{x+1}{3-x}}=\frac{4}{0^+}=+\infty$$

And so the range of $f$ is $[0,+\infty[$.

Does anyone knows other systematic ways of finding the range of a function without using the inverse? Thanks.

• It's not necessary to take a limit at $x = -1$. Dec 26, 2013 at 20:53

For differentiable functions you can use the characterization that the first derivative is zero at any extremal point in the interior of the intervall.

Here is another suggestion:

Use strictly monotonically increasing/decreasing functions to simplify the problem.

For those functions the range is determined by the behavior on the endpoints of the interval.

• Consider the term $\frac{x+1}{3-x}$. For the given domain it is easy to see that it has a zero in $-1$ and is positive otherwise. Thus at $-1$ there is a minimum.

• Next we use that $x \mapsto \frac{1}{x^2}$ is strictly monotonically decreasing for $x>0$ (because its composed of $x \mapsto x^2$ and $x\mapsto \frac{1}{x}$). So $f$ has an extremal point iff $\frac{1}{f(x)^2}$ has an extremal point. But $$\frac{1}{f(x)^2} = \frac{3-x}{x+1} = \frac{2}{x+1}-1$$ is clearly monotonically decreasing. Consequently $f$ is monotonically increasing and it has no extremal points in $(-1,3)$.

• So the limit towards the other end of the intervall determines the range.

Tip:

If $g\left(x\right):=f\left(3-4x\right)=\sqrt{\frac{1}{x}-1}$ then $g$ has the same range as $f$. It is simpler to handle.