The largest interval for which $x^{12}-x^9+x^4-x+1>0$

Question

The largest interval for which $x^{12}-x^9+x^4-x+1>0$

A) $-4<x<0$, B) $0<x<1$, C)$-100<x<100$, D) $-\infty<x<\infty$

My attempt is as follows:-

As coefficient of $x^{12}$ is positive, so given polynomial will be positive for $x\in(-\infty,a)\cup(b,c)\cdots\cdots (\lambda,\infty)$

This is just the rough estimation of set of values of $x$ for which polynomial can be positive.

But If we talk of largest interval, then it can happen that for all $x$ polynomial is positive, so in that case $-\infty<x<\infty$. But is this the correct approach because I am doubtful if the given polynomial can be factorized or not.

Answer 1

For $x\leq0$ it's obvious.

For $0<x\leq1$ we have: $$x^{12}-x^9+x^4-x+1=1-x+x^4(1-x^5)+x^{12}>0$$ and for $x>1$ we obtain:

$$x^{12}-x^9+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1>0.$$ Id est, the answer is D).

Answer 2

You can study the inequality: $$x^{12}-x^9+x^4-x>0$$ which is very simple. If we find that this inequality holds $\forall x \in R$ we have done. In other case, we have to compute some values of $x$ for which $f(x)<0$ and then detrmine whether $f(x)+1>0$ or $f(x)+1<0$.

Answer 3

It is clear that for $x=0$ the polynomial is positive, and hence it is positive on an open neighbourhood of $1$. This eliminates options A and B. It is also clear that $-100$ and $100$ are not zeros of the polynomial, excluding option C.

Alternatively, for negative $x$ the polynomial is positive. Then it can only be option D.

Answer 4

Quick answer: For $x<0$ all terms are positive, so the inequality is obeyed. Then it's option D.

Full answer

For $x\le 0$ all terms are non-negative, so the inequality is obeyed. For $x\ge 1$ we have $x^{12}\ge x^8$ and $x^4\ge x$, so once again all $x$ are good. What's left is $x\in(0,1)$. For that:$$x^12-x^8+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1=x(x^8+1)(x^3-1)+1$$ We know that $x^8+1$ is going to be in the interval $(1,2)$, and since $x^3-1$ is negative, we can just replace $x^8+1$ with $2$. If it works for this value, then all is fine.

So what's left is to calculate the minimum value of $x^4-x$: $$(x^4-x)'=4x^3-1=0$$ $$x_{MIN}=\frac 1{\sqrt[3]4}$$ Then $$x(x^3-1)|_{MIN}=-\frac 1{\sqrt[3]4}\frac 34$$
We multiply with the maximum value for $x^8+1$, which is $2$. The inequality is now $$1-\frac1{\sqrt[3]4}\frac 32>0$$Move the negative term to the right and cube: $$1\gt \frac{3^3}{4\cdot 2^3}=\frac{27}{32}$$ This is a true statement, so , when putting everything together we get that the inequality is obeyed for any $x\in\mathbb R$

Answer 5

The sequence

is a Sturm sequence for your polynomial. Their limits to $+\infty$ have signs $+,-,-,+,+,+,+,-,-,+$, which has $4$ sign changes. Their limits to $-\infty$ have signs $+,+,+,+,-,+,-,-,+,+$, which has also $4$ sign changes. Therefore, there are no real roots of that polynomial. Since its value at $x=0$ is $1$, which is positive, then all its values on the reals are positive too.

Computed with this calculator.

Answer 6

Let $f(x)=x^{12}-x^9+x^4-x+1.$

Let $x>1$, then $$f(x)= x^{12}-x^9+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1>1$$ $$\implies f(x) > 1$$ Next let $x<1$, then $$f(x)=x^{12}+x^4(1-x^5)+(1-x)>0.$$ $$\implies f(x)>0$$ Let $x=0$, then $f(0)=1.$ Hence $f(x) > 0, \forall x \in (-\infty,\infty)$