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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?

Shuster
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    This isn't actually injective at all. In particular, $f(0)=-\frac45=-.8$, $f(10)=\frac{26}{105}=0.247\dots$, and $f(-100)=-0.030\dots$, so $f(x)$ must hit $-0.5$, for example, at least twice. – boink Aug 27 '19 at 01:07
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    Welcome to Math Stack Exchange. As @boink commented, $f(x)=\dfrac{3x-4}{x^2+15}$ not injective: $f(0)=f\left(\dfrac{-15}4\right)$ – J. W. Tanner Aug 27 '19 at 01:10

5 Answers5

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Desmos showing the graph of this function is :enter image description here

The horizontal line near the $y$ value $-1$(for example) intersect twice the graph ,concluding this function is not one-one!

Chinnapparaj R
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You can show the non-injectivity of the function without finding specific values $x_0 \neq x_1$ with $f(x_0)=f(x_1)$ as follows:

  • $f(0) = -\frac{4}{5}$
  • $\lim_{x\to \color{blue}{-\infty}}f(x)= 0$
  • $\lim_{x\to \color{blue}{+\infty}}f(x)= 0$

Applying the intermediate-value property for continuous functions you immediately get the existence of $x_0 \in (\color{blue}{-\infty},0)$ and $x_1 \in (0,\color{blue}{+\infty})$ with $f(x_0) = f(x_1) = -\frac{1}{4}$. Done.

trancelocation
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When a function is injective, you have that $f(x_1) = f(x_2)$ implies in $x_1 = x_2$ for all elements in the domain of the function you are working with.

Take $f(0)=f\left(\dfrac{-15}4\right)$ as @J.W.Tanner said in the comments. In that case, you have $f(x_1) = f(x_2)$. However, $x_1$ is not equal to $x_2$.

A counterexample to the implication, since you have at least one element that makes it not happen.

$f(x_1) = f(x_2)$ does not imply in $x_1 = x_2$.

So, the function $f(x) = \frac{3x-4}{x^2+5}$ is not injective (one-to-one).

Daniel Sehn Colao
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$$f(x) = \frac{3x-4}{x^2+5}$$

$$f(0)=-\dfrac 45 = f\left(-\dfrac{15}{4}\right)$$

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 $k=0$

Steven Alexis Gregory
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By your work we obtain $$3x_1x_2(x_1-x_2)-15(x_1-x_2)-4(x_1^2-x_2^2)=0$$ or $$(x_1-x_2)(3x_1x_2-15-4(x_1+x_2))=0,$$ which gives also $$3x_1x_2-15-4(x_1+x_2)=0$$ for $x_1\neq x_2,$ which says that our function is not injective.

Michael Rozenberg
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