M1 2 math module 1 solutions to exercises y x 0 50

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sumption possibilities. M1-2 MATH MODULE 1: SOLUTIONS TO EXERCISES y x 0 (0, 50) Slope = –2 (25, 0) FIGURE FOR ANSWER M.1-2
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M1-3 MATH MODULE 1: SOLUTIONS TO EXERCISES ( 6, 4) (0, 4) (6, 4) ( 6, 0) (0, 0) (6, 0) x y f b d e a c x FIGURE FOR ANSWER M.1-3 3. These 6 functions make a “scalloped” pattern. The reason is that they all result from displacing the basic function ( y = x 2 ) upward by 4 units and/or left or right by 6 units. Notice, however, that the final , right-hand versions of the 6 equations do not immediately reveal that this is what is going on. It is sometimes necessary to pull equations apart or do some rough graphing to figure out what they are actually say- ing. [Do not be afraid of using “trial-and-error” on a question, if you don’t fully understand it initially. The best mathematicians use such techniques all the time!]
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M1-4 MATH MODULE 1: SOLUTIONS TO EXERCISES y y x y = x 3 y = x 3 x y = x y = x 1 3 1 3 FIGURE FOR ANSWER M.1-4 4. Both of these functions have an inverse function. The inverse function for y = x 3 is x = y 1/3 , while the inverse function for y = x 1/3 is x = y 3 . Notice that if you simply held a mirror to the black function and then rotated the mirror-image a quarter-turn clockwise, you would end up with the blue function. If this question posed any prob- lems, you should consult Module 8.
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5. In this question, you have actually created some “standard” demand curves, by implementing the ceteris paribus (“other determinants assumed unchanged”) assumption. When these other variables change, the “standard” demand curve shifts in the ways you have calculated, although the overall demand function has not in fact changed. (a) Given X D = 20 – P X + 0.5 P Y + 5 M , we simply substitute in the values for P Y and M to get X D = 20 – P X + 0.5(20) + 5(4) = 50 – P X , or P X = 50 – X D .
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