Find the minimum value of the average cost for the given cost function on the given intervals.
\[
C(x)=x^{3}+37 x+128
\]
(a) $1 \leq x \leq 10$
(b) $10 \leq x \leq 20$

Step 1 ：Find the derivative of the cost function \(C(x) = x^3 + 37x + 128\), which is \(C'(x) = 3x^2 + 37\)

Step 2 ：Set the derivative equal to zero and solve for \(x\): \(3x^2 + 37 = 0\) => \(3x^2 = -37\) => \(x^2 = -37/3\)

Step 3 ：Since \(x^2\) cannot be negative, there are no real solutions to this equation, and therefore no critical points

Step 4 ：For the interval 1 ≤ \(x\) ≤ 10, evaluate the function at the endpoints: \(C(1) = 1^3 + 37*1 + 128 = 166\) and \(C(10) = 10^3 + 37*10 + 128 = 1238\)

Step 5 ：\(\boxed{166}\) is the minimum value of the average cost on the interval 1 ≤ \(x\) ≤ 10

Step 6 ：For the interval 10 ≤ \(x\) ≤ 20, evaluate the function at the endpoints: \(C(10) = 10^3 + 37*10 + 128 = 1238\) and \(C(20) = 20^3 + 37*20 + 128 = 8488\)

Step 7 ：\(\boxed{1238}\) is the minimum value of the average cost on the interval 10 ≤ \(x\) ≤ 20