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Let $\psi : \mathbb R \to \mathbb R$ be a continuous function such that $\lim_{x\to+\infty}\psi(x) = +\infty$ and $\lim_{x\to-\infty}\psi(x) = -\infty$. If we define a function $G:\mathbb R \to \mathbb R$ such that $$G(x) = \frac{\psi(x)}{1+\psi^2(x)}$$ How can we evaluate whether $G$ has a maximum and a minimum?

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    Hint: What can you say about $G$ at $\pm \infty$? Now consider the extreme value theorem. – 2017-01-28
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    @JohnHughes Can I apply Rolle's theorem? As stated the function is not necessarily differentiable. – 2017-01-28
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    Ooops -- wrong theorem. :( – 2017-01-28
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    @JohnHughes Extreme value theorem applies to compact sets. – 2017-01-28
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    That's why I said "*consider* the extreme value theorem." You have to actually make a case that there's a compact set on which it applies. – 2017-01-28
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    @JohnHughes I agree, but I was making sure. – 2017-01-28

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Since $\psi$ is surjective, there an $a\in\Bbb R$ with $\psi(a)=1$, hence $G(a)=1/2$ Similarly, there is a $b\in\Bbb R$ with $\psi(b)=-1$, and $G(b)=-1/2$. Let $M>0$ be large so that $0 M$, and $m<0$ large (in the sense of absolute value) such that $0>G(x)>-1/2$ for $x

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We have that $\lim_{x \to - \infty} G(x) = 0^-$, $\lim_{x \to \infty} G(x) = 0^+$ so exists $M > 0$ such that $G(x) > 0$ for all $x > M$ and $G(x) < 0$ for all $x < - M$. How $G$ is continuous, by Bolzano theorem, exists $c \in [- (M + 1) , (M + 1)]$ such that $G(c) = 0$. By extreme value theorem, exists $d_1 \in [- (M + 1) , c]$ such that $G(d_1) = \min_{x \in \mathbb{R}} G(x)$ and exists $d_2 \in [c , M + 1]$ such that $G(d_2) = \max_{x \in \mathbb{R}} G(x)$.

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    What's $f(x)$ here? – 2017-01-28
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    Sorry $G$ I change. Thanks – 2017-01-28
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    I am used to $f$ instead of $G$. – 2017-01-28