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!

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