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I'm asked to determine if a function is surjective or not, and formally prove it. The function $$f:\mathbb{R}\rightarrow\mathbb{R}, f(x) = x^4+x^2$$ is not surjective (I'm prety sure),I know for a counter-example to use a negative number, but I'm just having trouble going around writing the proof.

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  • $\begingroup$ The title doesn't match the body of the question. Which one are you asking? The proposition holds true for even degree polynomials in general, but the proof is next to trivial if you only meant $f(x)=x^4+x^2\,$. $\endgroup$ – dxiv Feb 22 '17 at 7:47
  • $\begingroup$ I edited the title to be more specific $\endgroup$ – Alex Feb 22 '17 at 7:50
  • $\begingroup$ As a sum of two squares $f$ can't take negative values. $\endgroup$ – Michael Hoppe Feb 22 '17 at 8:56
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Your proof can either be directly or by contradiction.

Proof: Note that $f:\mathbb R \to \mathbb R$. Suppose $f$ is onto (a.k.a surjective). Then this would imply that (imput your negative number here) is mapped to, and thus $x^2(x^2+1)=$ (said negative number). This would imply that $x$ is (by via some standard arithmetic that you can do) not in $\mathbb R$ or $x\notin \mathbb R$. Which by the way is a contradiction.

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  1. $ \lim_{x \to \infty}f(x)=\lim_{x \to -\infty}= \infty$.

  2. $f(x) \ge 0$ for all $x$ and $f(0)=0$.

Since $f$ is continuous, we derive

$$f(\mathbb R)=[0,\infty) \ne \mathbb R.$$

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It is easier than you possibly thought. Note that the codomain is $\mathbb{R}$. So let $y < 0$. Note that $x^{2} + x^{4} \geq 0$ for all $x \in \mathbb{R}$. So there is no $x \in \mathbb{R}$, the domain of $f$, such that $f(x) = x^{2}+x^{4} = y$. By definition $f$ is not surjective.

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Note that $t^2+t+1$ has no roots in $\Bbb R$, since the discriminant is negative.

Now suppose that $f(x)=-1$ for some $x$, set $t=x^2$, then $t^2+t=-1$, but this means that $t^2+t+1$ has a real root.

Essentially what you want to do is to formally prove that $f(x)\neq -1$ for all $x$.

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