Is function $f(x,y)=x(4-y^2)$ injective / surjective? Sketch $f^{-1}[B]$, where $b = [0 ; +\infty)$. There is a fuction $f: \mathbb{R}^2 \implies \mathbb{R} $  given by:
$$f(x,y)=x(4-y^2)$$

*

*Is the function injective?

*Is the function surjective?

*find and sketch $f^{-1}[B]$, where $b = [0 ; +\infty)$
Here is an sketch:
sketch (I am not allowed to share images)
As we can see, function $f$ can be described as a linear function with all it's rotations besides those that would cover the blue part of cartesian coordinate system. The function gets most vertival for $y = 0$.
Therefore, I would say that the function is injective.
However, I am not sure if it's surjective since for $y = -2$ the function is constant and returns only the value $0$.
And finally, the last part. I know that $f^{-1}[B]$, for $b = [0 ; +\infty)$ is an counterimage. The problem is, that in that case I have 2 separate types of argements. I would say that the answer is:

*

*{$n \in (-2 ; 0)\cup(0;2)\}\cap $ {$x \in (0; +\infty)$}

*{$n \in (-\infty ; -2)\cup(2;\infty)\}\cap $ {$x \in (-\infty;0)$}

 A: For any $a \in \mathbb{R}$, $f(\frac{a}{4},0)=a$. So surjective.
$f(x,2)=0$ for any $x$. So not injective.
For the image of $f^{-1}[B]$, use case analysis method:

*

*$|y|>2 \longrightarrow$ $x \in (-\infty,0]$

*$|y|=2 \longrightarrow$ $x$ is any real number.

*$|y|<2 \longrightarrow$ $x \in [0,\infty)$
Here is the image

A: This function is not injective.  For all $x,y \in \Bbb{R}$, $f(x,y) = f(x,-y)$, so most points in the image have at least two preimages.  (In fact, most points have infinitely many preimages.)
This function is easily seen to be surjective.  Let $u \in \Bbb{R}$ be an arbitrary point in the codomain.  Then $f(u,\sqrt{3}) = u$.
For $f(x,y) \geq 0$: the product of two terms is nonnegative when both are nonnegative or both are nonpositive.  So,
$$  \{(x,y) \mid x \geq 0 \text{ and } 4 - y^2 \geq 0\} \\ \quad \cup \{(x,y) \mid x \leq 0 \text{ and } 4 - y^2 \leq 0\}  \text{.}  $$
Easy simplifications: \begin{align*}
& \{(x,y) \mid x \geq 0 \text{ and } 4 - y^2 \geq 0\}  \\
    &\quad {} \cup \{(x,y) \mid x \leq 0 \text{ and } 4 - y^2 \leq 0\}  \\
    &\quad {} = \{(x,y) \mid x \geq 0 \text{ and } 4 \geq y^2\}  \\
        &\qquad {} \cup \{(x,y) \mid x \leq 0 \text{ and } 4 \leq y^2\}  \\
    &\quad {} = \{(x,y) \mid x \geq 0 \text{ and } -2 \leq y \leq 2\}  \\
        &\qquad {} \cup \{(x,y) \mid x \leq 0 \text{ and } ( y \leq -2 \text{ or } 2 \leq y)\}
\end{align*}
And this is easy to plot.

