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Show that the function $x^4 – 3x^2 + x-10$ cannot have a root inside $(0,2)$. Please note that roots of $f'(x)$ cannot be found using a calculator. Attempted the question by calculating $f'(x)$ and assuming that at least one root of $f(x)$ exists in $(0,2)$. However had difficulty locating the points of maxima/minima, which cannot be done without finding the roots to $f'(x)$. Think there's another way to do it. Any suggestions?

Amzoti
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Completed
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4 Answers4

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You can show that in the domain $(0,2)$,

$$ x^4 - 3x^2 + x - 10 < -10 + 3x.$$

This is equivalent to

$$x (x-2) (x+1)^2 < 0. $$

Since $-10 + 3x < 0 $ in the domain $(0,2)$, it follows that the function is never 0 in the domain.

The RHS of the inequality is obtained by finding the linear function which satisfies the values at the end points. This helps us by ensuring that we know 2 of the roots.

Calvin Lin
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  • Cannot factorize without knowing the roots right? The motive here is to prove that the above function has no roots in the given domain. Even more, the inequality is not strictly '<'. – Completed Sep 02 '13 at 06:48
  • Is it possible to use MVT & Rolle's theorem to prove the above. Or using calculus? – Completed Sep 02 '13 at 06:55
  • @Completed Note that I'm factorizing another polynomial, namely $x^4 - 3x^2 - 2x$. From my construction, I know that $x=0, 2$ are roots, so I'm left with a quadratic polynomial. Also, the inequality does hold for $<$, since we are within the domain, i.e. not the endpoints. – Calvin Lin Sep 02 '13 at 07:40
  • Apologies I didn't note that. Any definite argument to show that $x^4−3x^2+x−10<−10+3x$. – Completed Sep 02 '13 at 10:34
  • @Completed Read the very next line ... – Calvin Lin Sep 02 '13 at 10:40
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Note that the function $g(x) = x^4-3x^2$ achieves maximum modulus in $[0,2]$ at $x=2$, with $|g(4)|=4$. Also note that in $(0,2)$ we have $$|x^4-3x^2+x-10|>10-|x|-|g(x)| > 8-|g(x)| \geq 4.$$

Jonathan Y.
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Calvin's method looks great [+1]. If you really want other ways, you could try

If $0 < x < 2$, then $x^3 < 8$ and $3x^2 + 10 > 2 \sqrt{30} x$ by AM-GM.

So $x^4 + x = x(x^3+1) < 9 x < 2 \sqrt{30} x < 3x^2 + 10$

Macavity
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Each term of a polynomial function $f:x\mapsto \sum_{k=0}^na_kx^k$ is monotonic in $x$ over any interval $[a,b]$ with $0\leqslant a\leqslant b$. Therefore $$f(x)\le \sum_{k=0}^n a_kc_k^k\quad(a\leqslant x\leqslant b),$$where $c_k=a$ if $a_k\leqslant 0$ and $c_k=b$ if $a_k>0.$ Now you can apply this to your example by checking this upper bound in each of (say) the intervals $[0,\sqrt2]$, $[\sqrt2,\sqrt3],$ and $[\sqrt3,2]$, which is easy to calculate.

John Bentin
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