Exam 3 | MA 22300 - Introductory Analysis I, Quizzes of Mathematics

Download Exam 3 | MA 22300 - Introductory Analysis I and more Quizzes Mathematics in PDF only on Docsity! TERM 1 Let f(x) = (2x + 1)3 (4 x)2 . (a). Find the intervals of increase and decrease of f(x). (b). If any exist, find and classify each critical point of f(x) as a relative minimum, relative maximum, or neither. DEFINITION 1 (a). First, f0(x) = 2(2x + 1)2(13 x)(4 x)3 . We then have 1/2, 4, 13 as criticalnumbers as this is when f0 is undefined or zero. We then see that f0 is positive,or increasing, when x < 1/2, 1/2 < x < 4, x > 13 and f0 is negative, ordecreasing, when 4 < x < 13.(b). We only have two points to consider: (1/2, 0) and (13, 243) as f is not definedwhen x = 4. Using either the second derivative test or a sign chart from theprevious part, we see that (1/2, 0) is neither and (13, 243) is a relative minimum TERM 2 The population of a certain small town between the years of 1995 and 2004 is given by P(t) = t3 3t2 24t + 247 where t is the number of years after 1995. At what time during this period was the population shrinking most rapidly? DEFINITION 2 Solution. We want when the population is shrinking most rapidly so we want theabsolute minimum of the rate of change of population. In other words, we want theabsolute minimum of P0(t) when 0 ​ t ​ 9. Now, P0(t) = 3t2 6t 24 and soP00(t) = 6t 6 = 0 =) t = 1. Then P0(0) = 24, P0(1) = 27, and P0(9) = 165.Therefore, the population is shrinking most rapidly when t = 1 or during 1996. TERM 3 Let f(x) = 3x5 + 20x4 15x 771. (a). Find the intervals on which the graph of f(x) is concave up or concave down. (b). Find any points of inflection. DEFINITION 3 Solution. (a). f0(x) = 15x4+80x315 and so f00(x) = 60x3+240x2 = 60x2(x+4) =0 =) x = 0,4. Since f00 is defined everywhere, we only need to look at thesetwo numbers. We then have that f00 > 0, or f is concave up, when 1 < x < 0and 0 < x. f00 < 0, or concave down, when x < 4.(b). Note that f00 changes sign about x = 4 but not x = 0. Therefore, the onlyinflection point is (4, 1337). TERM 4 Let f(x) = 8x2 x2 5x 14. Find all vertical and horizontal asymptotes, if any exist. DEFINITION 4 Solution. Note, f(x) = 8x2(x 7)(x + 2) so we have vertical asymptotes at x = 7and x = 2 as the denominator is zero there. We have a horizontal asymptote aty = limx!1 f(x) = 8. TERM 5 Find the absolute extrema of the function f(x) = x3 3x2 9x + 1 on 2 ​ x ​ 2. DEFINITION 5 Solution. First, f0(x) = 3x2 6x 9 = 3(x 3)(x + 1) = 0 =) x = 3, x = 1.Note that x = 3 is not in the interval we are considering, so we can ignore it. Wenow check the value of f at x = 1 and the endpoints x = 2, and x = 2. Thenf(2) = 1, f(1) = 6, and f(2) = 21. Therefore, f has its absolute max at (1, 6)and its absolute min at (2,21). TERM 6 Find dy dx if 4x xy2 = 1 y. DEFINITION 6 Differentiating both sides we obtain 4y2 2xydydx= dydxwhere we usedthe product rule for the xy2 term. Solving for dydxwe have dydx= 4 y22xy 1. TERM 7 Determine where the function f(x) = (2x + 1)3 (4 x)2 is increasing and decreasing DEFINITION 7 First,f0(x) = 6(2x + 1)2(4 x)2 + 2(4 x)(2x + 1)3(4 x)4= 2(2x + 1)2(4 x)[3(4 x) + (2x + 1)](4 x)4= 2(2x + 1)2(13 x)(4 x)3We then have that f0(x) = 0 or undefined when x = 12, 4, 13. Looking at the signs ofeach term we see that f0(x) > 0, or f is increasing, when x < 12, 12< x < 4, andx > 13. We also have that f0(x) < 0, or f is decreasing, when 4 < x < 13. TERM 8 The cost of manufacturing x cases of soda is C(x) = 2 5x2 p x + 230 dollars. The weekly production t weeks from now is given by x(t) = 25t + 125 cases. Find the rate at which cost will be changing with respect to time 3 weeks from now. Is it increasing or decreasing? DEFINITION 8 Solution. We want the derivative of C(x(t)) evaluated at t = 3. Using the chain ruleits derivative evaluated at 3 is C0(x(3))x0(3) ​ 4000 as x(3) = 200, x0(3) = 25, andC0(200) ​ 160. Alternatively, one can write out C(x(t)) explicitly and differentiate,in which case C(x(t)) = 25(25t + 125)2 p25t + 125 + 230 and so its derivative is20(25t + 125) 252p25t + 125. Plugging in t = 3 we again obtain approximately 4000.Thus, it is increasing. TERM 9 A cylindrical ice sculpture is currently melting. Its volume is currently decreasing at a rate of 18​ ft3 per hour and its height is decreasing at a rate of 1 ft per hour. Find the rate at which its radius is changing when its volume is 64​ ft3 and its height is 4 feet. (Hint: V = ​r2h) DEFINITION 9 Solution. We are given dvdt = 18​, dhdt = 1, V = 64​, h = 4 and we want to finddrdt . First, we find the radius, r. Plugging what were given in to the volume we have64​ = 4​r2 =) 16 = r2 =) r = 4. Now, differentiating, dVdt= 2​rhdrdt+ ​r2 dhdt.Plugging in what were given and solving this becomes 18​ = 2​(4)(4)drdt+​(4)2(1)and so 2​ = 32​drdt=)drdt= 116. TERM 10 Use calculus to estimate how much the function f(x) = x3 + 3x + 1 changes as x decreases from 2 to 1. DEFINITION 10 Solution. The estimation formula is ​f = f0(a)​x where ​x = b a and so in thiscase it is ​f = f0(2)(1). Differentiating, f0(x) = 3x2+3 and so f0(2) = 15. Pluggingthis back into the estimation formula we have ​f = 15.