What is the relation that is verified by the real solution of equation y= 4x^3-6x^2-24x+a? -26

First, we'll calculate the roots of the derivative of the
function. These roots are the critical values of the given
function.

y = f(x)

f'(x) =
12x^2 - 12x - 24

We'll put f'(x) =
0

12x^2 - 12x - 24 = 0

We'll
divide by 12:

x^2 - x - 2 =
0

The roots of the first derivative of the function are: x1
= -1 and x2 = 2.

Now, we'll calculate the local extremes of
the function:

f(-1) = 4*(-1) - 6 + 24 +
a

f(-1) = 14 + a

a + 14 =
0

a = -14

f(2) = 32 - 24 - 48
+ a

f(2) = -40 + a

a - 40 =
0

a = 40

We notice that for
values of a in the interval (-14 ; 40), 14 + a > 0 and  a - 40 <
0.

According to Rolle's
theorem:

x1 is in the interval (-infinite ;
-1)

x2 is in the interval (-1
; 2)

x3 is in the interval (2
; +infinite)