Find the smallest value of the function y = 3 cos x + 10x + 5 on the interval [0; 3p / 2].

1. Find the derivative of the function and examine it for monotonicity:
y = 3cosx + 10x + 5;
y ‘= -3sinx + 10 = -3sinx + 3 + 7 = 7 + 3 (1 – sinx) ≥ 7> 0.
The derivative is everywhere positive, the function increases on the set of real numbers.
2. The smallest value of the increasing function on the segment [0; 3π / 2] will be on its left border:
y = 3cosx + 10x + 5;
ymin = y (0) = 3cos0 + 10 * 0 + 5 = 3 * 1 + 5 = 8.
Answer: 8.