Calculus
posted by Wiz .
If the derivative can be thought of as a marginal revenue function for x units (in hundreds of items) sold, and the revenue for a company is given by the function.
R(x) = 30x^3  120x^2 + 500 for 0 _< x _< 100,
a. Sketch the graphs of the functions R(x) and R'(x) .
b. Find the number of units sold at which the marginal revenue begins to increase

marginal revenue begins to increase when f'' changes sign.
R'(x) = 90x^2  240x
R"(x) = 180x240 = 60(3x4)
So, marginal revenue begins to increase when x = 4/3 
r(x) = 30x^3  120x^2 + 500
r ‘ (x) = 90x^2  240x
0 = 90x^2  240x
0 = 3x^2  8x
0 = x(3x – 8)
X = 0 or x = 8/3
Or approx.
X = 3
Answer: x = 3
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