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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!

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  • $\begingroup$ 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. $\endgroup$ – Acccumulation Aug 22 '18 at 22:43
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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 $\mathbb{R}\to (-\infty,2)$.

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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.

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

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