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So I have to find an interval (in the real numbers) such that it contains all roots of the following function: $$f(x)=x^5+x^4+x^3+x^2+1$$
I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.
HINT: Determine how many real roots there are. If there is only one root, how big is the interval?
Let $f(x)=x^5+x^4+x^3+x^2+1$.
Hence, $f'(x)=x(5x^3+4x^2+3x+2)$ and since $(5x^3+4x^2+3x+2)'=15x^2+8x+3>0$, we see that the polynomial $5x^3+4x^2+3x+2$ has one real root $x_1$ and this root is negative.
Thus, $x_{min}=0$ and $x_{max}=x_1$.
But $f(0)>0$, which say that $f$ has an unique real root and since $$f(-1.25)f(-1.24)<0,$$ we get that this root placed in $(-1.25,-1.24)$.
Note that $$ f(x) = x^5+x^4+x^3+x^2+1 = (x + 1)(x^4 + x^2) + 1. $$
Now try answering some question about $f(x)$ for numbers that are relatively easy to work with:
Can $f(x)$ be zero if $x > 0$? Can $f(x)$ be zero if $x < 0$?
Can $f(x)$ be zero if $x > -1$? Can $f(x)$ be zero if $x < -1$?
Can $f(x)$ be zero if $x > -2$? Can $f(x)$ be zero if $x < -2$?
If you haven't found an interval by this time, you can try other numbers.
EDITED for correct function.
$f(x) = 1+x^2+x^3+x^4+x^5$ is positive for x>0.
So consider $x < 0$:
One can write $f(x) = (1+x^2)(1+x^3) + x^4$ which is positive at least for $(1+x^3) >0$, i.e. $x > -1$.
$f'(x) = 2x+3x^2+4x^3+5x^4 = x(2+3x)+x^3(4+5x)$ so this is positive at least for $x < -4/5 $ since then all terms are positive.
This means that $f(x) = 0$ can only happen at $x < -1$ and since $f(x)$ is rising there, we have that there can only be one real root, which is at $x < -1$.
Now this root can be further locked in. Observing e.g. that $f(-1.5) = -85/32 <0$, we can give an interval for the root at (-1.5 , -1). This can obviously be improved, knowing that there is only one real root and that f(x) is monotonously increasing in this interval.
Can you find a positive number $N$ such that $N^5>N^4+N^3+N^2+1$? If $|z|\ge N$ then one cannot have $z^5=-z^4-z^3-z^2-1$.
HINT: observe that
$$\lim_{x\to +\infty}f(x)=+\infty,\qquad\lim_{x\to -\infty}f(x)=-\infty$$
and
$$\lim_{x\to \pm\infty}\frac{4x^4}{x^5}=0$$
Thus exists some $\alpha>1$ such that $$|x^5|>|4x^4|>| x^4+x^3+x^2+1|$$
when $x>\alpha$ or $x<-\alpha$.
The common factoring formula $x^n-1=(x-1)(x^{n-1}+x^{n-2}+ \cdots + x^2+1)$ tells us that your $f(x)$ is equivalent to $g(x)=\dfrac {x^6-1}{x-1}$ as long as $x \not = 1$. Setting $g(x)=0$ yields $x^6-1=0$. Clearly the only real solutions to this are $\pm1$. But only $-1$ is a zero of $g$ and of $f$, so $f$ has only one real zero.
