Math 113 HW #5 Solutions


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1 Mat 3 HW #5 Solutions. Exercise.5.6. Suppose f is continuous on [, 5] and te only solutions of te equation f(x) = 6 are x = and x =. If f() = 8, explain wy f(3) > 6. Answer: Suppose we ad tat f(3) 6. Ten eiter f(3) = 6 or f(3) < 6. Te first is clearly impossible, because we know f(x) = 6 only wen x =,. On te oter and, if f(3) < 6, ten we ave tat f is continuous on [, 3] and f() = 8 > 6 > f(3). Terefore, by te Intermediate Value Teorem, tere exists a number c between and 3 suc tat f(c) = 6. However, tis is again impossible because c is between and 3 and so cannot be equal to or. Hence, since f(3) 6 is impossible, we ave to conclude tat f(3) > 6, as desired.. Exercise.6.6. Sketc te grap of an example of a function f tat satisfies all of te conditions f(x) =, x 0 + f(x) =, x 0 f(x) =, x f(x) =. x Answer: Exercise.6.8. Find te it y 3y 5y + y Answer: Dividing bot numerator and denominator by y, we get ( y 3y ) y (5y y + y) 3 y y 5 +, y
2 wic, using te Limit Law for quotients, is equal to ( ) y 3 y ) = y ( y Terefore,. Exercise.6.8. Find te it y 3y 5y + y = 3 5. cos x. x Answer: Because cos(nπ) = for any n and because cos((n + )π) = for any n, tis it cannot exist: no matter ow far out we go on te xaxis, cos x is still oscillating between and, so it never settles down to a it. 5. Exercise (a) A tank contains 5000 L of pure water. Brine tat contains 30 g of salt per liter of water is pumped into te tank at a rate of 5 L/min. Sow tat te concentration of salt after t minutes (in grams per liter) is C(t) = 30t 00 + t. Answer: Since te tank starts wit 5000 L of water and since 5 L/min of brine pours in, te number of liters of water in te tank is given by t. On te oter and, te tank starts wit no salt in it and, for eac liter of brine tat pours in, 30 g of salt pours in. Hence, te number of grams of salt in te tank is given by 5(30t). Terefore, te concentration of salt is given by C(t) = 5(30t) t = 5(30t) 5(00 + t) = 30t 00 + t. (b) Wat appens to te concentration as t? Answer: As t, te concentration of salt is given by 30t C(t) t t 00 + t. Dividing bot numerator and denominator by t yields t t t 30t (00 + t) t t + = 30. So eventually te concentration of salt in te tank approaces 30 g/l, wic is te same as te concentration of salt in te brine. In oter words, te brine eventually overpowers te pure water.
3 6. Exercise.7.. Sown are graps of te position functions of two runners, A and B, wo run a 00m race and finis in a tie. (a) Describe and compare ow te runners run te race Answer: A runs te race at a constant speed, never speeding up or slowing down. B accelerates trougout te race, starting out slower tan A and, by te end, running faster tan A. (b) At wat time is te distance between te runners te greatest? Answer: Based on te grap, it appears tat tey are furtest apart after 8 seconds, wen tey are approximately 30 meters apart. (c) At wat time do tey ave te same velocity? Answer: Te two graps appear to ave te same slope (i.e., velocity) 9 or 0 seconds into te race. 7. Exercise.7.0. Sketc te grap of a function g for wic g(0) = g (0) = 0, g ( ) =, g () = 3, g () =. Answer: 8. Exercise.7.. If g(x) = x 3, find g (0) and use it to find an equation of te tangent line to te curve y = x 3 at te point (0, ). Answer: Since g(0) =, te slope of te tangent line to te curve y = x 3 is given by g (0). Now, by definition, [ g g(0 + ) g(0) (0 + ) 3 ] [ 0 3] 3 3 (0) 0. Now, so long as 0, we ave tat 3 =. Terefore, since 0 as 0. 3 ( ) = 0 0 Tus, te tangent line is orizontal; te only orizontal line troug (0, ) is te line y =, so we see tat te equation of te tangent line is y =. 3
4 9. Exercise.7.3. Te it 6 + represents te derivative of some function f at some number a. State suc an f and a. Answer: If we let f(x) = 6 + x and coose a = 0, ten f f(0 + ) f(0) 6 + (0 + ) (a) , wic is te given it. Terefore, tis it represents f (0) for f(x) = 6 + x. 0. Exercise.7.8. Te quantity (in pounds) of a gourmet ground coffee tat is sold by a coffee company at a price of p dollars per pound is Q = f(p). (a) Wat is te meaning of te derivative f (8)? Wat are its units? Answer: Te derivative of a function always gives te cange in te output wit respect to te cange in te input. In tis case, since te input to te function is te price and te output is te quantity sold, te derivative gives te cange in quantity sold wit respect to a cange in te price. Tis is essentially wat economists call te price elasticity of demand. In particular, f (8) will give te expected cange in quantity sold if te price is canged from $8/lb. Te units are pounds per dollar. (b) Is f (8) positive or negative? Explain. Answer: f (8) will almost certainly be negative. Remember tat f f(8 + ) f(8) (8). If > 0, ten f(8+) represents te quantity of coffee sold for some price tat s greater tan $8/lb., wic we would certainly expect to be less tan f(8) (te quantity sold wen te price is exactly $8/lb.). Hence, te number f(8 + ) f(8) will be negative and so, since > 0, f(8 + ) f(8) < 0. On te oter and, wen < 0, te numerator inside te above it will be te difference between te quantity sold wen te price is less tan $8/lb. and te quantity sold wen te price is exactly $8/lb. Tis will almost certainly be a positive number, so again (since < 0) we ll ave f(8 + ) f(8) Hence, te expression inside te it will always be negative, so we would expect f (8) to be negative. < 0.
5 Colloquially, tis just says tat we would expect te quantity sold to go down wen te price goes up, and te quantity sold to go up wen te price goes down. 5
f(a + h) f(a) f (a) = lim
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