<s>[INST] Show that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{3x-4}{x^2+5}$ is injective I'm working on proving that $f: \Bbb{R} \rightarrow \Bbb{R}$ defined by $f(x) = \frac{3x-4}{x^2+5}$ is injective. However I'm stuck.
Assuming that $f(x_1) = f(x_2)$:
$$f(x_1) = f(x_2)$$
$$\frac{3x_1-4}{x_1^2+5} = \frac{3x_2-4}{x_2^2+5}$$
$$(3x_1-4)(x_2^2+5) = (3x_2-4)(x_1^2+5)$$
$$3x_1x_2^2 + 15x_1 - 4x_2^2 - 20 = 3x_2x_1^2 + 15x_2 - 4x_1^2 - 20$$
$$3x_1x_2^2 + 15x_1 - 4x_2^2 = 3x_2x_1^2 + 15x_2 - 4x_1^2$$
How do I get to $x_1 = x_2$ from here on out?
 [/INST] $$f(x) = \frac{3x-4}{x^2+5}$$
$$f(0)=-\dfrac 45 = f\left(-\dfrac{15}{4}\right)$$
You would have had an easier time solving 
\begin{align}
   \dfrac{3x-4}{x^2+5} &= k \\
   3x-4 &= kx^2 + 5k \\
   kx^2 - 3x +(5k+4) &= 0 \\
   x &= \dfrac{3 \pm \sqrt{9-16k-20k^2}}{2k}
\end{align}
So $x$ is double valued for all 
$$-\dfrac 25 - \dfrac{\sqrt{61}}{10} < k < -\dfrac 25 + \dfrac{\sqrt{61}}{10}$$
except it is single-valued at $k=0$
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