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. $10 \leq x \leq 20$ The minimum value of the average cost over the interval $1 \leq x \leq 10$ is 85.0 . (Round to the nearest tenth as needed.) The minimum value of the average cost over the interval $10 \leq x \leq 20$ is (Round to the nearest tenth as needed.)

Step 1 ：The cost function is given by \(C(x) = x^3 + 37x + 128\).

Step 2 ：The average cost function is given by \(A(x) = \frac{C(x)}{x} = x^2 + 37 + \frac{128}{x}\).

Step 3 ：To find the minimum value of the average cost function, we need to find the derivative of the average cost function and set it equal to zero.

Step 4 ：The derivative of the average cost function is given by \(A'(x) = 2x - \frac{128}{x^2}\).

Step 5 ：Setting the derivative equal to zero gives us \(2x - \frac{128}{x^2} = 0\).

Step 6 ：Solving this equation for \(x\) gives us \(x = \sqrt[3]{64} = 4\).

Step 7 ：However, \(x = 4\) is not in the interval \(1 \leq x \leq 10\) or \(10 \leq x \leq 20\).

Step 8 ：Therefore, the minimum value of the average cost function on the interval \(1 \leq x \leq 10\) is \(A(1) = 1^2 + 37 + \frac{128}{1} = 166\).

Step 9 ：The minimum value of the average cost function on the interval \(10 \leq x \leq 20\) is \(A(10) = 10^2 + 37 + \frac{128}{10} = 165\).

Step 10 ：Therefore, the minimum value of the average cost over the interval \(1 \leq x \leq 10\) is \(\boxed{166}\) and the minimum value of the average cost over the interval \(10 \leq x \leq 20\) is \(\boxed{165}\).