A more rigorous way to show that $x^5 - 3x = 1$ has at least $3$ roots in $\Bbb R$ 
Given an equation:
  $$
x^5 - 3x = 1
$$
  Show that:
  
  
*
  
*It has at least $1$ root on $(1, 2)$;
  
*It has at least $3$ roots on $\Bbb R$

I've started with considering a function $f(x)$ for $x\in [1, 2]$:
$$
f(x) = x^5 - 3x - 1
$$
Then calculating its value on the left and right sides of the closed interval yields:
$$
f(1) = -3\\
f(2) = 25
$$
Applying the Intermediate Value Theorem yields that there exists a point for $x_0 \in [1, 2]$ such that $f(x_0) = 0$. Which means that indeed at least one root exists.
However, for the second part of the question if we consider $f(x)$ for $x \in \Bbb R$, the only way I see is to try and guess the intervals where the function changes its sign and then apply IVT again. Consider for example $f(x)$ for $x \in \{-2, -1, 1, 2\}$.
I see how derivatives could be to the rescue here, the problem is that I'm not allowed to use derivatives.
Is there a rigorous way to prove what's stated without guessing and without using derivatives? Thank you!
 A: Let us use Descartes' sign Rule
$f(x)=x^5-3x-1=0$, the number of sign changes in $f(x)$ is one and number of sigm changes in $f(-x)$ are two. So as per Descarte's rule this equation can have atmost one real positive and atmost two real negative roots. So Descarte's rules permits at most three real root which can be located as : $f(-\infty)<0, f(-1)=+1, f(0)=-1$ means there is at least one real root in $(-\infty, -1)$ and at least one in $(-1,0)$. Next $f(1)=-3, f(2)=+25$.
So one real positive roots is in $(1,2)$. In all this equation has three real roots two negative and one positive.
A: Graphical method can help.
$$x^5-3x=1 \stackrel{x\ne 0}{\iff} x^4-3=\frac1x$$
Sketch:
$\hspace{4cm}$
Note: You can quickly jump to the intervals of interest and compare the values and increase/decrease of the functions.
A: (1)$f(x)=x^5-3x-1$ has one sign change in its coefficients so at most $1$ positive root. Also, $f(1).f(2)<0$ assures exactly one positive root lying in $(1,2)$.
(2) $f(-x)=-x^5+3x-1$ has two sign changes in its coefficients so at most $2$ negative roots OR more precisely $0$ or $2$ negative roots (possibility of exactly $1$ negative root is discarded due to pairwise occurrence of remaining complex conjugate roots). Now $f(-1).f(0)<0$ is sufficient to assure the existence of exactly $2$ negative roots of $f(x)$.
A: The coefficient $-3$ in the middle is dominant enough to allow to consider the two binomial expressions $x^5-3x$ and $-3x-1$ from the sides of the Newton polygon as separate approximations for large and small roots, with their real roots at $\pm\sqrt[4]3$ and $-\frac13$. To get a more precise connection to the given polynomial, consider the product of the binomials
$$
(x^4-3)(x+\tfrac13)=x^5+\tfrac13x^4-3x-1.
$$
This is close to the given polynomial $x^5-3x-1$ so that in moving from one to the other the roots do not change much. The pairwise distance between the $3$ real roots and the $2$ imaginary roots $\pm i\sqrt[4]3$ is large enough that the nature of these roots will not change under the perturbation to the given polynomial. To verify this more or less heuristic consideration, compute the values of the polynomial at enclosing intervals around $-\frac43, -\frac13,\frac43$, for instance $[-\frac32,-1]$, $[-\frac12,0]$, $[1,\frac32]$.

A bit more certainty and general applicability one gets from computing the root radius. Recall that for $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ a radius enclosing all the roots is obtained as
$$
R=1+\max_{k<n}|a_k|\text{ or } R=\max(1,|a_{n-1}|+...+|a_1|+|a_0|)
$$
which gives here $R=4$ in both cases. As there is a large degree gap, scaling the polynomial might help reduce the radius, set $x=2y$, then 
$$
f(2y)=32y^5-6y-1=32(y^5-\tfrac3{16}y-\tfrac1{32})
$$
which results in the bounds $R=2+\frac3{16}$ or $R=2$ for $x$. Now that the range of the roots is sufficiently restrained, computing values at $x=-2,-1,0,1,2$ is now a well-founded strategy. Counting sign changes as per Descartes then confirms that there are exactly two negative roots.

A more precise guessing can be performed using derivatives.
$f'(x)=5x^4-3$ has 2 real roots at $\pm\sqrt[4]{\frac35}$, thus $f(x)$ has local extrema there. Check the function value at the local maximum $-\sqrt[4]{\frac35}$, for simplicity check at the close-by point $x=-1$:
$$
f(-1)=-1+3-1=1>0.
$$
Together with $f(0)=-1$ and $f(-\infty)=-\infty$ one can conclude for the existence of $2$ negative roots. 
There can be no more roots, as then by Rolle the derivative would have to have more than $2$ real roots.
A: Quick and dirty without calculus. Plugging in some small integer values in $x^5 -3x -1$ (always a good place to start on this kind of problem) shows that the sign changes between $-2$ and $-1$, then again between $-1$ and $0$, then again before the values grow to $\infty$. That guarantees at least three real roots.
A: This is inefficient and it may not always work but
$x^5 - 3x-1 = x(x^4-3)-1$.
If we can find $a<b$ but that $a(a^4 - 3) \ge 1$ and $b$ so that $b(b^4 -3) \le 1$, then we will done. (Why?[*]) 
If we can find an $a$ where $a < 0; k= a^4 - 3 < 0$ then $ak > 0$. And if  $|a| \ge 1$ and $|k| \ge 1$ then $ak \ge 1$.  And that will do for our $a$. 
(Note: this is sufficient but much stronger and much more than necessary.)
That's easy enough.  $k = a^4 - 3 \le -1$ implies $a^4 \le 2$ implies $|a| \le \sqrt[4]{2}$ so we just need $-\sqrt[4]{2} \le a \le -1$ for our $a$.
Likewise  if we have $b >0$ but $j = b^4-3 < 0$. Because $bj < 0$ and $bj - 1< -1 < 0$.    (And again sufficient, albeit much stronger than necessary.
So $b^4 -3  <0$ so $b^4 < 3$ so $|b| < \sqrt[4]{3}$ so any $b; 0 < b < \sqrt[4]{3}$ will do.
[*]Thus  we have $\lim_{x\to -\infty} x^5 - 3x -1 = -\infty$ so there are plenty of $n < a$ so that $n^5 -3n -1 < 0$.  And $a(a^4-3) -1 > 0$ so there is a root between $n$ and $a$.  And $b(b^4-3) -1 < -2 < 0$ so there is a root between $a$ and $b$.  And $\lim_{x\to \infty} x^5 - 3x -1 = \infty$ so there are plenty of $m> b$ so that $m^5 -3m -1 > 0$ so there is a root between $b$ and $m$.
But that was inefficient and required some educated guesses.
