016A Sep 9 Homework SolutionJae-young Park∗September 6, 2008• 0.1 *18 If f(x) =x2x2−1, find f(12), f (−12), and f(a + 1)Solution f(12) =(1/2)2(1/2)2−1=1/41/4−1=1/4−3/4= −13f(−12) =(−1/2)2(−1/2)2−1=1/41/4−1=1/4−3/4= −13f(a + 1) =(a+1)2(a+1)2−1=a2+2a+1a2+2a+1−1=a2+2a+1a2+2a• 0.1 *54 Compute f(1), f(2), f(3) for the following funtionf(x) =½1/x for 1 ≤ x ≤ 2x2for 2 < xSolution For x = 1, since 1 ≤ x = 1 ≤ 2, we get f(1) =1x|x=1= 1.For x = 2, since 1 ≤ x = 2 ≤ 2, we have f(2) =1x|x=2=12. For x = 3,since 2 < x = 3, f(3) = x2|x=3= 32= 9• 0.2 *14 A line has x-intercept (−500, 0) and y-intercept (0, 60). De-termine K and V where the equation of the line is y =KVx +1V.Solution The slope of the line isslope =0 − 60(−500) − 0=−60−500= 0.12and y intercept is 60. So we haveKV= 0.121V= 60∗jaypark at m a t h . b e r k e l e y . e d u. GSI for 16A 101,104,1051Solving this equation, we getV =160K =1500• 0.2 *32 Sketch the graph.Solution Go to the end of the solution.• 0.3 *14 Express f(t) − h(t) as rational function where f(x) =xx−2and h(x) =x+13x−1Solution Note that f(t) =tt−2notxx−2f(t) − h(t) =tt − 2−t + 13t − 1=t(3t − 1) − (t + 1)(t − 2)(t − 2)(3t − 1)The numerator can be simplified.t(3t − 1) − (t + 1)(t − 2) = 3t2− t − (t2− t − 2) = 2t2+ 2So, we getf(t) − h(t) =2t2+ 2(t − 2)(3t − 1)=2t2+ 23t2− 7t + 2• 0.3 *30 Calculate f(h(x)) where f(x) = x6and h(x) = x3− 5x2+ 1Solutionf(h(x)) = f(x3− 5x2+ 1) = (x3− 5x2+ 1)6• 0.3 *34 If g(t) = t3+ 5, findg(t+h)−g(t)hand simplifySolution To calculate g(t + h), substitute t + h for t and simplify.g(t + h) − g(t)h=[(t + h)3+ 5] − [t3+ 5]h5’s in the numerator are canceled and expanding (t + h)3, we getg(t+h)−g(t)h=t3+3t2h+3th2+h3−t3h=3t2h+3th2+h3h= 3t2+ 3th + h22• 0.3 *36 During the first 1/2 hour, the employees of a machine shopprepare the work area for the day’s work. After that, they turn out10 precision machine parts per hour, so that the output after t hoursis f(t) machine parts, where f(t) = 10(t −12) = 10t − 5,12≤ t ≤ 8.The total cost of producing x machine parts is C(x) dollars, whereC(x) = .1x2+ 25x + 200.Solution(a) The total cost of producing x machine parts is C(x)dollars. So substituting f(t), which is a output after t hour, for x, weget the desired one.C(f(t)) = .1(10t − 5)2+ 25(10t − 5) + 200= .1(100t2− 100t + 25) + 250t − 125 + 200 = 10t2+ 240t + 77.5(b) Now we have a function C(f(t)) which is a cost after t hours ofoperation, so substituting 4 for t, we get the answer.C(f(4)) = 10(4)2+ 240 ∗ (4) + 77.5 = 160 + 960 + 77.5 = 1197.5So the cost after 4 hours is 1197.5 dollars• Let f (x) = 1+(1/(1+(1/(1+(1/x))))). Express f (a2+2) as a rationalfunction (quotient of two polynomials)Solution f(x) = 1 +11+11+1x. First, simplify this f(x)f(x) = 1 +11 +11+xx= 1 +11 +xx+1= 1 +12x+1x+1= 1 +x + 12x + 1=2 + 3x1 + 2xNow put x = a2+ 2.f(a2+ a) =2 + 3(a2+ 2)1 + 2(a2+ 2)=3a2+ 82a2+ 5• 0.4 *10 Using the quadratic formula, Solve z2−√2z −54= 0.Solution z doesn’t matter. Using the quadratic formula above,since a = 1, b = −√2, c = −54, we get the following answersz =√2 ±q(−√2)2− 4 ∗ 1 ∗ (−5/4)2 ∗ 1=√2 ±√72.So the roots are z =√2+√72,√2−√72.3• 0.4 *24 Factor f(x) = 16x + 6x2− x3.Solution First factor −xf(x) = −x(x2− 6x − 16)Now we factor x2− 6x − 16. If you use the quadratic formula,x =6 ±p36 − 4 ∗ 1 ∗ (−16)2= −2, 8So, we get f(x) = −x(x + 2)(x − 8). Or you can try to find pair ofnumbers (c, d) such that cd = −16 and c + d = −6. Factoring −16,you can find (2, −8) or (−8, 2).• 0.4 *30 Find the points of the intersection of the pairs of two curvesy =12x3− 2x2y = 2xSolution We have to solve12x3−2x2= 2x. Multiplying both sidesby 2 and adding −4x, we getx3− 4x2− 4x = x(x2− 4x − 4)Using the quadratic formula, we get the solution of the x2−4x−4 = 0.So, we find three solutions,x = 0, 2 + 2√2, 2 − 2√2And these are the x coordinates of the intersection points. Since theyare on the line y = 2x, we get the points of the intersection(0, 0) (2 − 2√2, 4 − 4√2) (2 + 2√2, 4 + 4√2)• 0.4 *36 Solveg the following equation1 =5x+6x2Solution First note that x 6= 0. Multiplying x2to both sides,x2= 5x + 6 → x2− 5x − 6 = 0Factoring, we get x2− 5x − 6 = (x − 6)(x + 1) = 0 So, x = −1, 6. Ifyou plug these back, you can check these are solutions.4• 0.5 *18 Compute the number 163/4Solution Apply the exponential laws. Note that 16 = 24.163/4= (24)3/4= 24×34= 23= 8• 0.5 *36 Find 20.55.5Solution Note 20 = 4 × 5. First use (ab)x= axbx20.55.5= (4 · 5).55.5= 4.55.55.5Now use axay= ax+y.4.55.55.5= 4.55.5+.5= 4.55Now use (ax)y= axy( 4 = 22).4.55 = (22).55 = 2 × 5 = 10• 0.5 *46 Simplify (x3y6)1/3Solution(x3y6)1/3= (x3)1/3(y6)1/3= x3·13y6·13= xy2• 0.5 *74 Fill in the following blank.rxy−ryx=√xy( )Solutionrxy−ryx=rxyy2−rxyy2If b > 0, we havepab=√a√b.rxyy2−rxyy2=√xypy2−√xy√x2=√xy(1py2−1√x2)So we haverxy−ryx=√xy(1|y|−1|x|)5• 0.5 *94 Assume that a couple invests 4000 dollars each year for fouryears in an investment that earnss 8% compounded annually. Whatwill be the value of the investment be 8 years after the first amount isinvested?Solution 8 years after the first amount is invested, the first invest-ment 4000 dollars is worth 4000(1 + 0.08)8dollars. Another three4000 dollars investments would be worth 4000(1 + 0.08)7,4000(1 +0.08)6,4000(1 + 0.08)5, respectively. So the value of the investmentis4000(1 + 0.08)8+ 4000(1 + 0.08)7+ 4000(1 + 0.08)6+ 4000(1 + 0.08)5It’s approximately 26484 dollars.• 0.6 *20 Suppose the cylinder in Exercise 6 has a volume 54π cubicinches. Find the surface area of the cylinder.Solution The cylinder in Exercise 6 has height h and the radiush2. So the volume is π(h2)2h = πh34Since the volume is 54π, we get thefollowing eqautionV = πh34= 54π → h = 6The area of the top and bottom each is π32= 9π. The area of the sideis (2πr)h = πh2= 36π. So the surface area is 2 × 9π + 36π = 54π• 0.6 *22 A college student earns income by typing term papers on acomputer. The student charges 4 dollars per page for her work andthe mothly cost is given by C(x) = .10x + 75 when she type x pagesfor a month.Solution (a) What is the profit if she types 100 pages?The revenue is given by R(x) = 4x. The profit function P (x) =R(x) − C(x).