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Is the below true or false?

If $f(x)$ is a non-constant real polynomial for which $g(x)=f(x^2)+1$ has a real root then $f(x)$ also has a real root?

If $f(x)=\sum_{n=0}^m a_nx^n$ with $\deg(f)=m$, then $f(x^2)=\sum_{n=0}^m a_nx^{2n}$ , $\deg(g)=\deg(f(x^2))=2m$ ($g(x)$ has only even powers) and if $x_0$ the real root of $g(x)$ then: $g(x_0)=0 \Rightarrow f(x_0^2)+1=0$, but I don't have anything else. What do I miss?

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    There's a $g$ missing in the first problem statement. – 2017-02-09
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    You may assume $m=\deg(f)$ is even. – 2017-02-09
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    Where's the $g$ missing? – 2017-02-09
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    If we restrict $f$ to be a polynomial with *positive* leading coefficient, then it'll be true. From $f(x_0^2) + 1=0$, we have a point $x_0^2 \in \Bbb R$ so that its function value is *negative*. As $x \to +\infty$, $a_m x^m$ will dominate $f(x)$, so it'll tend to $+\infty$. Use the Intermediate Value Theorem to show the existence of a real root of $f$. – 2017-02-09
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    I understand your concept: but can we apply this theorem on an open (or unbounded better) interval, since in $[x_0^2,\infty]$ you have a negative and positive value of $f(x)$ respectively, which means that since $f(x)$ is continuous on that interval (being a polynomial) there will be $y_0$ such that: $f(y_0)=0$? – 2017-02-09
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    This theorem actually applies to [a close and bounded interval](https://en.wikipedia.org/wiki/Intermediate_value_theorem#Theorem). – 2017-02-09
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    That I know, that's why I asked how you are going to use it to prove that $f$ has a root in your case! – 2017-02-09

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FALSE

Counterexample To The Claim: $$f(x)=-x^2-\frac{1}{2} \implies g(x)=f(x^2)+1=-x^4+\frac{1}{2}$$

$g(x)$ has a real root, but $f(x)$ does not.

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    Nice counterexample! – 2017-02-09
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    @JohnZobolas Yeah, but as GNU supporter said, your claim works if it has a positive leading coefficienet. – 2017-02-09