Problem 3: Consider 𝑓(𝑥) = 𝑥^2 (10 − 𝑥)^2/3 .

Question

a) Find the derivative of f. (Make it “visually appealing” by writing it as a single fraction, so find a common denominator.) b) Using the Calculus we have learned this quarter, find all the critical points of 𝑓 b) Using the Calculus we have learned this quarter, find the absolute min and the absolute max of 𝑓 on the interval [2, 10.5]. *Simply graphing and labelling will not earn you credit.*

Answer 1

a) f(x) = x²(10-x)^2/3

f '(x) = (2/3)x²/(10-x)^1/3+2x(10-x)^2/3 = (-8x²+60x)/(3(10-x)^2/3) = -(4/3)x(2x²-15)/(10-x)^2/3)

b)

This appears to be a test question. I can give you the process.

Find the critical points, where the derivative is equal to zero or undefined within the given interval. Calculate f(x) for those values and select the maximum and minimum values from that list of values.

Answer 2

This appears to be a test question. Consequently, it would not be ethical for me to work the problem. However, it is not difficult. This is the multiplication of two expressions. x² and (10 - x)^2/3, so you would use the product rule. Make sure to use the chain rule on (10 - x)^2/3. This will get you an answer in the form of a/b + c/d which you will need to get a common denominator to put in a single expression (as the problem says).

To find the critical points, set the numerator of the derivative developed equal to zero and solve.

To find the absolute max and min, check the endpoints on the interval and any critical points located on the interval.