Proving that $f(x)=x^2-6x-40$ is injective for $f: [3, \infty) \rightarrow [-49, \infty)$ without using calculus

Question

I'm trying to prove that $f(x)=x^2-6x-40$ is injective for $f: [3, \infty) \rightarrow [-49, \infty)$. Note that I cannot use calculus.

I tried letting $f(a)=f(b)$ and I arrived at $a^2-6a=b^2-6b$. Then I tried to find a solution for $a$ in terms of $b$ and a solution for $b$ in terms of $a$ and I got $a=\frac{6 \pm \sqrt{36+4b^2-24b}}{2}=3 \pm \sqrt{9+b^2-6b}$ and $b=\frac{6 \pm \sqrt{36+4a^2-24a}}{2}=3 \pm \sqrt{9+a^2-6a}$. If I tried to equate to show that $a=b$ I basically arrived at where I started from. Then I tried the approach of seeing what happens if $a>b$ and $a<b$ but I couldn't get far with that either since for some numbers $a^2<6a$ and for some numbers $a^2>6a$. I'm wondering what the best method to prove that $f$ is injecitve is.

Answer 1

First complete the square: $$ x^2 - 6x -40 = (x - 3)^2 - 49. $$ Now, if $f(a) = f(b)$ you'd get $$ (a - 3)^2 - 49 = (b - 3)^2 - 49. $$

Answer 2

You've got $b=3 \pm \sqrt{9+a^2-6a}=3\pm\sqrt{(a-3)^2}$

Since $a \ge 3$,

$$b=3\pm(a-3)=a$$ or $$b=6-a$$

Since $b \ge 3$, and $a \ge 3$, we conclude that it's only possible when $$b=a$$

Answer 3

Complete the square:

$y=(x-3)^2 -49.$

This is a normal parabola with vertex at

$(3,-49)$, which is the minimum.

$3 \le x,$ $x \in \mathbb{R}.$

Let $z: = x-3$, then

$Y=z^2 -49$, $0 \le z$, $z \in \mathbb{R}.$

This function $Y(z)$ is strictly monotonically increasing:

$z_1 \lt z_2$ $\rightarrow :$

$(z_1)^2 \lt (z_2)^2$, $\rightarrow: $

$Y_1 \lt Y_2.$

Hence injective.

Now revert to $x$.