Finding the Pre-image $\ f^{-1}([0,+\infty))\ $

Question

I am trying to find the pre-image of $f$, $\ f^{-1}([0,+\infty))\ $, where $f:\mathbb{R^2}\rightarrow\mathbb{R}$ and $f(x,y)=y(x-y).$

From a previous post, I was told to solve $f(x,y)\geq0$, but when I try to do this, I get $$ y\geq0 $$ and $$ y\leq x $$

This does not match the answer provided in the book, which also states that $$ y\leq 0 $$ $$ y\leq x $$

Why is this the case? It seems like a trivial question, but I don't understand.

Answer 1

It all goes back to the same thing : the product of two numbers is non-negative if and only if both of them have the same sign (or any one is zero).

Therefore, if $f(x,y) \geq 0$ then $y(x-y) \geq 0$, which means that the product of $y$ and $x-y$ is non-negative. So :

• Either $y$ is non-negative and $x - y$ is non-negative i.e. $y \geq 0, x \geq y$.

• Or, $y$ is non-positive and $x-y$ is non-positive i.e. $y \leq 0, x \leq y$.

Consequently, $f^{-1}([0,\infty)) = \{(x,y) : 0 \geq y \geq x\} \cup \{(x,y) : 0 \leq y \leq x\}$.

Answer 2

Remember that the resilt is positive also when the things you're multiplying are negative.

We want solve $y(x-y) \ge 0$. This happens $\iff (y \ge 0$ and $x-y \ge 0)$ or $(y \le 0$ and $x-y \le 0)$

From the first case we get $y \ge 0$ and $x \ge y$.

From the second case we get $y \le 0 $ and $x \le y$

Answer 3

The pre-image of $[0,+\infty)$ is defined as:

$$\{(x,y)\in\mathbb{R}^2\,\colon y(x-y)\geq 0\}.$$

Working on this expression we have:

$$y(x-y)\geq 0\iff xy-y^2\geq 0\iff y^2\leq xy.$$

If $y\geq 0$, we need $x\geq y$.

If $y<0$, we need $x\leq y$.