Module Two Discussion Question: Solve the problem below. For your initial post in Brightspace, copy the description of your cylinder in the box below and then enter your solution to all three parts (parts a, b, and c) of the problem. To copy the description of your cylinder, highlighting and using "copy" from here in Mobius and then using "paste" into Brightspace should work. Hint: This is similar to Question 63 in Section 5.7 of our textbook. We covered this material in the Module One section "Inverses and Radical Functions" within the Reading and Participation Activities. You can check some of your answers to parts b and c to make sure that you are on the right track. The height of the cylinder is 6 inches. We'll be analyzing the surface area of a round cylinder - in other words the amount of material needed to "make a can". A cylinder (round can) has a circular base and a circular top with vertical sides in between. Let r be the radius of the top of the can and let be the height. The surface area of the cylinder, A. is A=2x²+2πrh (it's two circles for the top and bottom plus a rolled up rectangle for the side). r-radius Areas ² Circumference 2x h-height Area = h(2) Part a: Assume that the height of your cylinder is 6 inches. Consider A as a function of r, so we can write that as A (r) = 2x² + 12 πr. What is the domain of A (r)? In other words, for which values of r is A (r) defined? Part b: Continue to assume that the height of your cylinder is 6 inches. Write the radius as a function of A. This is the inverse function to A (r), i.e to turn A as a function of into. as a function of A.