What is the best way to solve this high school exercise?

Question

Can you share with me how would you best solve this exersise to a high school student?

Show that $f(x)=x^2-6x+2$ , $x\in(-\infty,3]$ is $1-1$ and find its inverse.

Answer 1

Hint: To prove the function is 1-1 use the definition

$$ f(x_1)=f(x_2) \implies x_1 = x_2 .$$

Answer 2

Complete the square: $$f(x)=(x-3)^2-7$$

This is a parabola, pointing upward, with vertex at $(3,-7)$. You can graph it to supply the "show" portion. Then,

$$y=(x-3)^2-7$$ $$y+7=(x-3)^2$$ $$\sqrt{y+7}=|x-3|=3-x$$ (since $x<3$) $$x=3-\sqrt{y+7}$$ So $f(x)=3-\sqrt{x+7}$ is the inverse.

Answer 3

In addition to what vadim123 has written, by showing $f(x)$ is one-to-one in that domain, assume there are two $u, v\in(-\infty, 3]$ such that

$$\begin{align*} f(u) =& f(v)\\ u^2 - 6u + 2 =& v^2 - 6v + 2\\ u^2 - v^2 =& 6u - 6v\\ (u-v)(u+v-6) =& 0 \end{align*}$$

So either $u=v$, or $u+v = 6$. For the second case, since $u\le 3$ from the domain, $$v=6-u\ge6-3=3$$ this makes $v=3$ and hence $u=6-v = 3=v$. Anyway $f(u)=f(v)$ implies $u=v$.