First Principles AI

Solution

Let $g:\mathbb{R}\to\mathbb{R}$ be defined by:

g(x)=x^2-3x

Assume $g(x_1)=g(x_2)$ :

x_1^2-3x_1=x_2^2-3x_2

\begin{aligned} (x_1^2-x_2^2)-3(x_1-x_2)&=0\\ (x_1-x_2)(x_1+x_2)-3(x_1-x_2)&=0\\ (x_1-x_2)(x_1+x_2-3)&=0 \end{aligned}

So either $x_1=x_2$ or $x_1+x_2=3$ .

For example, $x_1=0$ and $x_2=3$ give:

g(0)=0,\quad g(3)=0

with $0\ne 3$ , so $g$ is not injective.

Let $y\in\mathbb{R}$ and solve $x^2-3x=y$ :

x^2-3x-y=0

This has real solutions only if the discriminant is non-negative:

\Delta=(-3)^2-4(1)(-y)=9+4y\ge 0\Rightarrow y\ge -\frac{9}{4}

So values $y<-\frac{9}{4}$ are not achieved, hence $g$ is not surjective.

\boxed{g\ \text{is neither injective nor surjective}}