# What is the relation that is verified by the real solution of equation y= 4x^3-6x^2-24x+a? -26<a<40

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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)**