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$$P(x)=ax^3+bx^2+cx+d; |P(x)| <1 \text{ for all } |x| <1$$ Prove that $$|P(x)| ≤|4x^3-3x|\text{ for all } |x| >1$$

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    are you sure the statement has no typo? I strongly believe the condition is $|x|\leq 1$ and $|x|>1$, respectively. Otherwise this problem reduces to a trivial problem. – 2017-01-15
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    @UchihaItachi You've changed the question from a quadratic to a cubic. Please make sure you know what you are asking as this looks poorly thought out. – 2017-01-15
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    can you help me :(( no this is the true question, it must'nt be wrong, my friends also say that too :(( – 2017-01-15
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    @UchihaItachi I am 99% certain that the statement has $|P(x)|\leq1$ for all $|x|\leq 1$, not $x<1$. Please check again. – 2017-01-15
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    yeah you right, sorry :(( – 2017-01-15

1 Answers 1

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Given $|P(x)|<1\ \forall\ x<1\ \ \ \ \ $ -----(I)

If $a\ne0$ then $\displaystyle\lim_{x\to-\infty}|P(x)|=\infty$, this violates (I), hence $a=0$.

Similarly for $b$ and $c$,

if $b\ne0$ then $\displaystyle\lim_{x\to-\infty}|P(x)|=\infty$, this violates (I), hence $b=0$,

if $c\ne0$ then $\displaystyle\lim_{x\to-\infty}|P(x)|=\infty$, this violates (I), hence $c=0$

The above implies that $-1

Also $4x^3-3x>1\ \forall\ x>1.$

Hence, $$|P(x)| ≤|4x^3-3x|\ \forall\ x >1$$

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    $P(x) = \dfrac{x^2+x}{100}$ satisfies the given condition but neither $a$ nor $b$ is zero. – 2017-01-15
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    @dezdichado It doesn't satisfy $|P(x)|<1$ for all $x<1$. – 2017-01-15
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    you are right, I read the condition as $|x|<1$. – 2017-01-15
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    thanks for your answer, but in this case I've already solved it. Sorry :(( I've edited the question. Many appologise :(( – 2017-01-15
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    sorry but I've made a mistake about the problem :(( please forgive me and help me just one more time :(( – 2017-01-15