How can you show that $f$ is one-to-one without horizontal line test? $f(x) = \sqrt{4x^2 + 1}$ for $x \in [0, \infty)$
I know that typically I can just do the horizontal line test, but I'm confused about how one would do this algebraically.
 A: Algebraic method: for one-one functions, show that $f(x_1) = f(x_2) \implies x_1 = x_2$:
\begin{align}
\sqrt{4(x_1)^2 + 1} &= \sqrt{4(x_2)^2 + 1} \\
4(x_1)^2 + 1 &= 4(x_2)^2 + 1 \\
4(x_1)^2 &= 4(x_2)^2 \\
(x_1)^2 &= (x_2)^2 \\
x_1 &= \pm \sqrt{(x_2)^2} \\
&= \pm x_2.
\end{align}
But because $x \geq 0$, we reject the negative root.
So $x_1 = x_2$, thus the function is injective.
A: *

*The function $x\mapsto \sqrt{x}: [0,\infty) \to [0,\infty)$ is one-to-one

*The function $x\mapsto 4x+1: [0,\infty) \to [0,\infty)$ is one-to-one

*The function $x\mapsto x^2: [0,\infty) \to [0,\infty)$ is one-to-one

Call these functions $f_1, f_2, f_3$. Then your function $f(x)=\sqrt{4x^2+1}$ is equal to
$$f(x) = f_1(f_2(f_3(x)))$$
Can you take it from here?
A: Hint: you can just show that the function is monotone, then this suffices for the function to be injective.
Hint 2: you have to show that the function doesn't change from positive to negative or to negative to positive, then the function will be monotone.
A: A kind of horizontal line test would be this:
$f(a)=f(b)$ iff $\sqrt{4a^2+1} = \sqrt{4b^2+1}$ iff $4a^2+1 = 4b^2+1$ iff $4a^2 = 4b^2$ iff $a^2=b^2$.
The latter means $a=b$ or $a=-b$. But the domain is nonnegative and so $a=b$.
