Surjectivity and injectivity
I need to show the injective and surjective for $f:\mathbb R^2 \longrightarrow \mathbb R$ where $f(x,y)=5xe^y$
For injective $f(0,0)=f(0,1)$ but $(0,0) \neq (0,1)$.
For surjective i must show that the function covers the codomain so that every value from $\mathbb R^2$ must have an exit value such that $f(x,y)=z$ right? How does that work?
Thanks!
$\endgroup$ 12 Answers
$\begingroup$For surjectivity, you could consider the ordered pair $(\frac{1}{5}z, 0)$ for a given $z \in \mathbb{R}$ as $f(\frac{1}{5}z, 0) = 5(\frac{1}{5}z)(e^0) = z$. We know $(\frac{1}{5}z, 0)$ is in the reals since $z$ is a real number, so $\frac{1}{5}z$ will also be real.
$\endgroup$ $\begingroup$For surjectivity you have to show that for any $z \in \def\R{\mathbf R}\R$ there is $(x,y) \in \R^2$ such that $f(x,y) = z$.
Hint. To do so, use that you (hopefully) know that $\exp \colon \R \to (0,\infty)$ is bijective. If for example $z > 0$, then we can choose $x = \frac 15$. Which leaves us with $$ f\left(\frac 15, y\right) = e^y \stackrel != z $$ So $y = \textbf ?$ will do.
For $z < 0$ you can use almost the same trick with another $x$-value. This leaves you with $z = 0$, but this is easy.
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