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 $\sqrt{145}<15$, $x^2>0$, 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:
$$ \left(-\infty, -\sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15+\sqrt{145}}{10}}, -\sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15+\sqrt{145}}{10}}, \infty\right) . $$
