Range and inverse of $f(x)=\sqrt{4x^2 + 1}$ on $[0,\infty)$

Question

Consider the function $f(x) = \sqrt{4x^2 + 1}$ for $x \in [0,\infty)$. Denote the range of $f$ by $Y$. There is a function $g$ with domain $Y$ so that $g(f(x)) = x$ for every $x \in [0, \infty)$ and $f(g(y)) = y$ for every $y \in Y$. Find an expression for $g(y)$

Also, how can I go about finding the range of $f$?

Any pointers would be much appreciated :).

Answer 1

Note that if $x \in [0,\infty)$ then $x \ge 0$.

So $x^2 \ge 0$. And $4x^2 \ge 0$ and $4x^2 + 1 \ge 1$ and $\sqrt{4x^2 + 1} \ge 1$.

So $Y \subset [1,\infty)$.

The only issue is can any $y \in [1,\infty)$ be in the range of $f$?

SO here are three questions for you to think about

1. If $y = \sqrt{4x^2 + 1}$ and $y \ge 1$ is the any limit on how big $y$ can be?
2. If we can have $y$ be as small as $1$ and we can have $y$ be as large as we like (can we? see question 1) does that mean we can have $y$ be any value in between? Is $f(x) = \sqrt{4x^2 + 1}$ continuous?
3. If we have any $k \in [1,\infty)$ can we always find an $x$ so that $k = \sqrt{4x^2 + 1}$? How would we know? Are there any $k$ *without solutions?

If after you answer all those questions, if every $k \ge 1$ will have an $x$ so that $\sqrt{4x^2 + 1} =k$ is possible, then that would mean the range of $f$ is $[1,\infty)$.

Is it?

Answer 2

Such a $g$ is know as the inverse of $f$, usually denoted $f^{-1}$. To find it, set $x=f(y)$ and rearrange to find $y$ in terms of $x$.

The domain of $f$ ($^*f^{-1}$) is equal to the range of $f^{-1}$ ($^*f$)