How do you know that the equation has three real roots between -2 and -1, one?

Question

EXAMPLE 1. Determine the number and situation of the real roots of
the equation x^5 - 3 x - 1 = 0.

The Sturm chain is

y = x^5 - 3 x - 1,

y' = 5 x^4 - 3,

y'' = 12 x + 5,

y''' = 1.

where y'= dy/dx

The signs of f for x = -2, -1, 0, +1, +2 are

-----------------------
x | f0 | f1 | f2 | f3
-----------------------
-2 | - | + | - | +
-1 | + | + | - | +
0 | - | - | + | +
+1 | - | + | + | +
+2 | + | + | + | +
-----------------------

The equation thus has three real roots: one between -2 and -1, one
between -1 and 0, one between +1 and +2. The other two roots are complex.

Question:
How do you know that the equation has three real roots between -2 and -1, one
between -1 and 0, one between +1 and +2???
Please explain!!

Answer 1

Let f(x) = x ^ 5 -3x - 1
Then f(-2) = (-2) ^ 5 - 3(-2) - 1 = -32 + 6 -1 = -27
f(-1) = - 1 + 3 -1 = +1
Since f is a polynomial and and therefore continuous, and since it changes from a negative value to a positive value as x goes from -2 to - 1, it must be 0, somewhere between - 2 and -1.
Similarly, since f(-1) = +1 and f(0) = - 1, the function must be 0 between -1 and + 1., and since
f(1) = -1 and f(2) = +25, there is another 0 value between 1 and 2.