Find the smallest value of the function y = 4 ^ x ^ 2 + 6x + 11

1. Let’s calculate the derivative of the function and find the critical point:

y = 4x ^ 2 + 6x + 11;

y ‘= 8x + 6;

8x + 6 = 0;

8x = -6;

x = -6/8 = -3/4, critical point.

2. Signs of the derivative in the intervals:

a) x ∈ (-∞; -3/4);

y ‘(- 1) = 8 * (-1) + 6 = -8 + 6 = -2 <0, the function decreases;

b) x ∈ (-3/4; ∞);

y ‘(0) = 8 * 0 + 6 = 6> 0, the function is increasing.

At the point x = -3/4, the function goes from decreasing to increasing, which means that this is the minimum point at which the function takes the smallest value:

y (min) = y (-3/4) = 4 * (-3/4) ^ 2 + 6 * (-3/4) + 11 = 4 * 9/16 – 18/4 + 11 = 9/4 – 9/2 + 11 = 44/4 – 9/4 = 35/4.

Answer: 35/4.