Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.

By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.
But I wonder that there is some logical ways we can find intervals that contain roots without guessing?
Given $f(x)=x^5-5x^3+4x-1$, differentiate first:
$$ f'(x) = 5x^4 - 15x^2 +4 . $$
Setting the derivative equal to $0$, we have
$$ 0 = 5x^4 - 15x^2 +4 . $$
This is quadratic in $x^2$, so we can use the quadratic formula:
$$ x^2 = \frac{15 \pm \sqrt{145}}{10} . $$
Since $\sqrt{145}&lt;15$, $x^2&gt;0$, so our solutions are
$$ x = \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}} . $$
So, we have four distinct local extrema for $f$. If there are five distinct roots, each must be on one of the following intervals:
$$ \left(-\infty, -\sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15+\sqrt{145}}{10}}, -\sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15+\sqrt{145}}{10}}, \infty\right) . $$
I would start with the rational root theorem, which fails but gets you values at $x=\pm1$.
Because the coefficients are small, the roots are all close to $0$.  Checking values at the integers close to $0$ is a good approach.  Once you go from $-3$ to $+3$ you have found one root in $(2,3)$.  You should be able to convince yourself there are not roots outside $(-3,3)$ because the $x^5$ term dominates.  Now checking at half integers is a good next step and finds the roots.  There was no guessing involved.  The real question comes if you don't find the roots with the half integral evaluations.  Do you trust the problem setter to guarantee there are five roots?  If so, trying quarters is the next logical step.  Or do you worry the question is wrong?  I have no advice here.