Group HW 8 - Julie Cprek Christie Donahue Jesse LaRochelle...

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Julie Cprek Homework 8 11/9/2006 Christie Donahue Jesse LaRochelle Evan Wladis 4/3 36. A) Why h’(x) is never negative on this interval: When a graph of the derivative is positive, the graph of the original function is increasing. When the graph of the derivative is negative, the graph of the original function is decreasing. There is a critical point of the graph of h’(x) at x=-1 since the graph of its derivative changes from negative to positive. This means there is a local minimum at that point. If the local minimum of the graph of h’(x) is at 0, there can be no point less than 0 on the graph. Thus, the graph must always be positive. An example of a possible h’(x) graph can be seen below in blue. The graph of h’’(x) is also plotted in black. B) Why h(x) has a global maximum at x=1: The graph is concave up for (-1,1] since the second derivative (the graph of concavity) is positive and the graph is concave down for [-2, -1) for the same reason. There is a point of inflection at x=-1 where the concavity switches from negative to positive. We know the slope of the graph of h(x) is 0 at x=-1 because h’(-1)=0. Knowing that the graph is concave until x=-1 we know that the slopes are the graph are increasing at a decreasing rate until they reach 0 at x=-1. Thus, the values of h(x) must be lesser to the left of x=-1. Since we know that the concavity is positive after x=-1, the graph must be concave up and the slopes must be increasing at an increasing rate. Since the h-values to the left of x=-1 are less than 2, the h-value at x=-1 is 2, and the h-values are always increasing to the right of x=-1, the right endpoint must be the global maximum because it is impossible for any other point to have a greater h- value than it. Similarly, h(x) has a global minimum at x=-2 because it is impossible for any other point on the graph to have a lesser h-value than it. C) A possible graph of h(x): We have established in parts a and b that the graph of h(x): is concave down for [-2,- 1); has a point of inflection at x=-1; is concave up for (-1, 1]; always has a positive slope (except at x=-1 where the slope is 0); the graph has a global minimum at x=-2 and a global maximum at x=1. A sample of the graph is shown below.
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Julie Cprek Homework 8 11/9/2006 Christie Donahue Jesse LaRochelle Evan Wladis 4/4 16. a) The cruise line charges $1000 per passenger for the first 100 passengers who make reservations. This means that, for the first 100 passengers, the cruise line has a revenue of 1000q where q is the number of passengers. If there are exactly 100 passengers, the revenue will be equal to 1000(100)=$100,000. Beyond 100 passengers, the cruise line chares $5 less per passenger up to 250 passengers which is the maximum capacity of the ship. This means that, for 101 passengers, the cruise line will charge $995 per passenger so the revenue will be 995(101)=$100,495. We are to find the number of passengers that maximizes revenue. To do this, we need to make an equation for revenue of the cruise line. Since there are two conditions to the equation, it must be a piecewise function.
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This note was uploaded on 04/02/2008 for the course MATH 115 taught by Professor Blakelock during the Fall '08 term at University of Michigan.

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Group HW 8 - Julie Cprek Christie Donahue Jesse LaRochelle...

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