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!
$\endgroup$ 12 Answers
$\begingroup$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 $\mathbb{R}\to (-\infty,2)$.
$\endgroup$ $\begingroup$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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