Find the smallest value of the function y = x3 + 12×2 + 36x + 3 on the segment [-5; -0.5].

| August 7, 2021 | education

Y = X ^ 3 + 12 * X ^ 2 + 36 * X + 3.

Y ‘(X) = 3 * X ^ 2 + 24 * X + 36 = X ^ 2 + 8 * X + 12.

Let us equate the derivative to zero and determine the critical points.

Solve the quadratic equation X ^ 2 + 8 * X + 12 = 0.

D = b ^ 2 – 4 * a * c = 8 ^ 2 – 4 * 1 * 12 = 64 – 48 = 16.

X1 = (-8 – √16) / 2 * 1 = (-8 – 4) / 2 = -12/2 = -6.

X2 = (-8 + √16) / 2 * 1 = (-8 + 4) / 2 = -4/2 = -2.

Point -2 belongs to the segment [-5; -0.5].

Let us determine the sign of the derivative at the point X = -1.

Y ‘(X) = 1 – 8 + 12 = 5. (More than zero, the function increases).

Let us determine the values of the function at points -5 and -0.5.

Y (-5) = -125 + 12 * 25 + 36 * (-5) + 3 = -2.

Y (-0.5) = -0.125 + 12 * 0.25 + 36 * (-0.5) + 3 = -12.125.

Answer: The smallest value in the segment is -12.125.



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