# Finding the range of a function without using inverse

So I'm fairly close to beginner level in calculus and have usually used the inverse of a function to find its range however I'm not sure what to do when dealing with this particular function. $$h(t) = \frac{t}{\sqrt{2-t}}$$ I found the domain to be $(-\infty, 2)$ but when I attempt to use the inverse to find the range, it ends up a mess because of the different powers of t. $$y = \frac{t}{\sqrt{2-t}}$$ $$\Rightarrow t = \frac{y}{\sqrt{2-y}}$$ $$\Rightarrow t^2 = \frac{y^2}{2-y}$$ $$\Rightarrow t^2(2-y) = y^2$$ $$\Rightarrow 2t^2-t^2y = y^2$$...

Maybe it's because I'm a beginner but I'm unsure where to go from here. Sorry if it's a really basic/easy question but I'd really like to learn how to deal with these types of questions. Any help would be appreciated!

• If you write it as $y^2+t^2y-2t^2=0$, you have the quadratic equation, with $a=1$, $b=t^2$, and $c=-2t^2$. However, you have to be careful about squaring both sides, especially when you're trying to find possible values like this. Only invertible transformations are guaranteed to preserve solution sets. Aug 22 '18 at 22:43

We have that $h(t)$ is a continuos function defined for $t<2$ and

$$\lim_{t \to -\infty} h(t)=-\infty$$

$$\lim_{t \to 2^-} h(t)=\infty$$

therefore by IVT the range is $\mathbb{R}$.

Morover we have

$$h'(t)=\frac{4-t}{2\sqrt{(2-t)^3}}>0$$

therefore $h(t)$ is also injective and the inverse exists from $\mathbb{R}\to (-\infty,2)$.

In fact, you don’t need to find $h$ inverse to find $h$ range. $h$ is a continuous map defined on $(-\infty ,2)$. Moreover, you have

$\lim\limits_{t \to 2^-} h(t)= \infty$ and $\lim\limits_{t \to -\infty} h(t)= -\infty$. Therefore the range of $h$ is whole $\mathbb R$ using the Intermediate Value Theorem.

• I like this answer, as it shows that you don't need inverses to answer such questions! Lots of functions don't have inverses. Aug 22 '18 at 19:20