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I have a function $f(x)$ and I want to prove that it is bijective. The function states

let $f (x) : \mathbb R \rightarrow \mathbb R$ be defined by $f (x) = -x^4$.

I understand that I need to show that it is injective and surjective but I don't understand how I would prove this for a function. I also know the definitions of injection and surjection.

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    $\begingroup$ this function is not injective nor surjective. $\endgroup$ Dec 3 '13 at 12:23
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    $\begingroup$ $f(1)=f(-1)$ and $f(x)<0$ for every $x\in \Bbb R$ $\endgroup$
    – Haha
    Dec 3 '13 at 12:24
  • $\begingroup$ Perhaps you wanted $f(x) = \mathrm{sign}(x)\cdot x^4$? $\endgroup$
    – dtldarek
    Dec 3 '13 at 13:52
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$f$ is injective if $x \neq y$ implies that $f(x) \neq f(y)$. $f$ is surjective if, for any $z$ in the co-domain (in this case $\mathbb{R})$, there is a point $x$ in the domain (again, $\mathbb{R}$), such that $f(x) = z$.

In this case, check that $f$ never attains a strictly positive value (for instance, $f(x) \neq 1$ for all $x\in \mathbb{R})$, so $f$ is not surjective.

Also, $f$ is not injective, because $1 \neq -1$, but $f(1) = f(-1)$

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$f : \mathbb{R} \to \mathbb{R}, f(x) = -x^4$ is neither injective nor surjective. Consider the following:

Take $y = f(2) = -16$. However, $f(2) = f(-2) = 16$, but $2 \neq -2$, which contradicts the definition of an injection. You can easily test for injectivity by plotting the graph (as you have a simple function) and using the horizontal line test.

The interval $(0, +\infty)$ is in the codomain $\mathbb{R}$, but there are no elements in the domain whose images fall into that interval. In other words, $-x^4$ is always in $(-\infty, 0]$ and never in $(0, +\infty)$. From here it follows that $f$ is not a surjection either.

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