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At the end of my class today, there was a proof that involved using the fact that given $f(x)=3+4x-x^2-x^4$

$f(x)\leqslant 0 $ if $|x|>2 $

this seems obvious but I was wondering if anyone could show me a more formal way of proving it?

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    Hint: Compute $f'$ to see where $f$ is increasing or decreasing. – 2017-02-19
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    f '(x) = 4-2x-4x^3 only has one real root – 2017-02-19
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    Therefore $f$ is monotone in any interval that does not contain that root, you just need to show that the intervals $(2,\infty)$ and $(-\infty, -2)$ do not contain that root (use Bolzano). – 2017-02-19
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    ah I see, and how can I use this to show there exists c s.t. f(c)>f(x) for all x – 2017-02-19
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    Once you've shown that $f$ is decreasing in $(2,\infty)$, that implies $f(x)\leq f(2)$, then just show that $f(2)<0$ and you are all done (after doing the same for $-2$). – 2017-02-19
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    You can use Bolzano to show that there is a root in the interval $[0,2]$, if you already have shown that there is only one real root, that means that there are no roots in the two intervals we are interested on. – 2017-02-19

2 Answers 2

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For $x\geq2$ we have $$x^4+x^2-4x-3=x^4-2x^3+2x^3-4x^2+3x^2-6x+2x-4+1=$$ $$=(x-2)(x^3+2x^2+3x+2)+1>0$$ For $x\leq-2$ we obtain: $$x^4+x^2-4x-3=x^4+3x^3+4x^2-4x^3-16x^2-16x+13x^2+52x+52-40x-80+25=$$ $$=(x+2)^2(x^2-4x+13)-40(x+2)+25>0$$

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Hint.   $|x| \gt 2 \implies x^4 \gt 16 \implies f(x) \lt 3+4x-x^2-16 = -x^2 +4x -13\,$ where the latter is a quadratic with no real roots.


P.S.  Not that it's needed to answer the question as posed, but it can be noted that the quartic actually factors into a product of two quadratics:

$$ \begin{align} f(x)=3+4x-x^2-x^4 & = (1+x-x^2)(3+x+x^2) \end{align} $$

The only real roots are $\cfrac{1 \pm \sqrt{5}}{2}\,$, so $\,f(x) \lt 0\,$ for $\,x \not \in \left[\cfrac{1 - \sqrt{5}}{2}, \cfrac{1 + \sqrt{5}}{2}\right]$.