Math Warm-Up for Exam 1 Name: Solutions

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1 Disclaimer: Tese review problems do not represent te exact questions tat will appear te exam. Tis is just a warm-up to elp you begin studying. It is your responsibility to review te omework problems, webwork problems, quizzes, workseets, and material from te text tat we ave covered up to tis point. 1. Find te equation of te line passing troug te points (5, 2) and (3, 1). Identify tis line s slope and y-intercept. Te slope is y 1 y 2 = 1 2 = 3/2. Ten we plug one point into y = ( 3/2)x + b to x 1 x find tat 1 = 9/2 + b and b = 11/2. So te equation of te line is y = ( 3/2)x + (11/2), te slope is 3/2, and te y-intercept is 11/2. 2. Determine te slope and y-intercept of te line given by te equation 3y 2x = 2. We rewrite tis equation as y = (2/3)x (2/3) to see tat te slope is (2/3) and te y-intercept is 2/3. 3. Write an equation tat represents te following function: te time, t, tat it takes to cross a 100 meter beac on foot is proportional to te square root of te dept, d, of te sand. t = k d for some constant k. 4. A truck rental company carges $100 to rent a truck for 24 ours and $300 to rent a truck for 120 ours (five days). (a) Find a linear model for te price, P, as a function of te number of ours rented, t. Slope = ( )/(120 24) = 200/96 = 25/12. Now 100 = (25/12)(24) + b so b = 50 and our model is P (t) = (25/12)t (b) Wat is te slope of te line in part (a)? Wat are te units of te slope? 25/ dollars per our (c) Wat is te value of te y-intercept found in part (a)? y-intercept? 50 dollars Wat are te units of te 5. A town as a population of 200 at time t = 0. In eac of te following cases, write a formula for te population P as a function of year t. (a) Te population increases by ten people eac year. P (t) = t (b) Te population increases by 4% eac year. P (t) = 200( ) t 6. Te Plague of Justinian, te first recorded epidemic of te bubonic plague, reduced te population of Constantinople by 40% between te years 541 and 700. (a) Assuming tat te population of Constantinople was 500,000 in te year 541, write a formula for an exponential function P (t) tat represents te population in terms of years, t, since 541 CE.

2 P 0 = 500, 000a t. Te population in 700 was (1 0.40)(500, 000) = 300, 000 and = 159 so 300, 000 = 500, 000a Solving for a gives a = 159 3/ , so P (t) = 500, 000( ) t (b) In wat year did Constantinople s population drop below 400,000? 400, 000 = 500, 000( ) t means t = ln(4/5)/ ln( ) 69, so te population dropped below 400,000 in te year = 610 CE. 7. If you write te function P = 34e 5t in te form P = P 0 a t, ten wat are P 0 and a? Is tis function an example of exponential growt or exponential decay? P 0 = 34 and a = e 5, exponential growt since a > 1 8. In te early 1960s, radioactive strontium-90 was released during atmosperic testing of nuclear weapons and got into te bones of people alive at te time. (a) If te alf-life of strontium-90 is 31 years, wat fraction of te strontium-90 absorbed in 1965 remained in people s bones in 2011? = 46. Fraction of strontium-90 is given by F (t) = a t, 1/2 = a 3 1 so a Now F (46) = (b) How many years does it take until only 5 percent of te original amount absorbed remains? 0.05 = t so t = ln(0.05)/ ln( ) 134 years. 9. For te functions f(x) = x 2 and g(x) = 2e x, find f(g(2)), f(g( 1)), g(f(t)), and f(a)g(a). f(g(2)) = (2e 2 ) 2 = 4e 4 f(g( 1)) = (2e 1 ) 2 = (2/e) 2 = 4/(e 2 ) g(f(t)) = 2e (t2 ) f(a)g(a) = a 2 2e a 10. If f(x) = 5 1 x, find f(t + 2), f(t2 ), and (f(t)) 2 3. f(t + 2) = 5 1 5t + 10 = t + 2 t t + 2 = 5t + 9 t + 2 f(t 2 ) = 5 1 t 2 (f(t)) 2 3 = (5 1 t )2 3 = t = 22 t2 t + 1 (Tis migt even simplify more.) t2 11. Are te following functions invertible? (a) f(n) is te number of pairs of soes sold by one soe store on te n-t day of probably not (b) f(p) is te number of sandwices needed to feed a lunctime population of p people. yes (c) f(d) is te temperature at a distance d west of Boulder. no

3 12. If f and g are bot odd functions wit f(2) = 3 and g(3) = 5, find g(f( 2)). g(f( 2)) = g( f(2)) = g( 3) = g(3) = 5 = Solve te following equations for x using logs. (a) 34 = 20(1.23) x 1.7 = 34/20 = 1.23 x ln(1.7) = x ln(1.23) x = ln(1.7)/ ln(1.23) 2.56 (b) 4(3) x = 2 x+2 ln(4 3 x ) = ln(2 x+2 ) ln(4) + x ln(3) = (x + 2) ln(2) x(ln(3) ln(2)) = 2 ln(2) ln(4) x ln(3/2) = ln(4) ln(4) = 0 x = 0 (c) 2e 10x = 5e 3x 5/2 = (e 10x )/(e 3x ) 5/2 = e 7x ln(5/2) = 7x x = ln(5/2)/ (d) Re x = Ca x (Assume R, a, and C are positive constants.) R/C = (a x )/(e x ) = (a/e) x ln(r/c) = x ln(a/e) x = ln(r/c)/ ln(a/e) ( 14. Consider te function y = sin x π ). 2 (a) Wat is te amplitude? A = 40 (b) Wat is te period? P = 2π B = 2π 1 = 2π (c) Wat is te pase sift? Has te grap been sifted to te left or to te rigt? C = π to te rigt 2 (d) Wat is te frequency? F = 1 P = 1 2π (e) Wen x = 0, wat is te value of y? ( y = sin 0 π ) = ( 1) = = (f) Wat is te largest value attained by y? Max value occurs wen sin = 1, so y = (1) = 240.

4 15. Te grap of a function f(x) looks exactly like te grap of y = sin(x) except tat it as a period of 7π. Write a formula for f(x). B = 2π P = 2π 7π = 2. Amplitude does not cange, no pase sift, no vertical sift. So ( ) 7 2 f(x) = sin 7 x 16. Find te formula for te grap below, given tat it is a polynomial, tat all zeros of te polynomial are sown, tat te exponents of eac of te zeros are te least possible, and tat it passes troug te point (-2,10). Te zeros occur at x = 3, 0, 4 and te one at x = 3 is squared. So f(x) = k(x+3) 2 (x)(x 4). Now we plug in our point to find k: 10 = k(1) 2 ( 2)( 6) so k = 10/12 = 5/6. So f(x) = (5/6)x(x + 3) 2 (x 4). 17. According to Car and Driver, a Volkswagen Jetta going 70 mp requires 193 feet to stop. Assuming tat te stopping distance is proportional to te square of te velocity, find te stopping distance required by a Jetta going at 15 mp and at 45 mp. D = kv ft = 193/5280 mi 193/5280 = k(70) 2 k D(15) = ( )(15) mi = 8.9 ft D(45) = ( )(45) mi = 79.8 ft 18. Find k so tat te function is continuous on any interval. { kx wen x 2 f(x) = 3x 2 wen x < 2 We need k(2) = 3(2) 2, so k = Te figure below gives a grap of f(x). Evaluate te following limits:

5 (a) lim f(x) = 4 x 2 (b) lim f(x) = 2 + x 2 (c) lim f(x) DNE x 2 (d) lim f(x) = 0 x Te figure below is te grap of a function f(x). Consider te slope of te grap of f at eac of te points A troug F, and put tese slopes in order from least to greatest. Also decide at eac point (A troug F ) weter te grap is concave up or concave down. At x = A and x = D concave up, at x = B, x = C, x = E, and x = F, concave down. Looks to me like f (F ) < f (C) < f (E) < f (A) < f (B) < f (D). 21. Estimate (4 + ) 2 16 lim using your calculator. Ten evaluate te limit algebraically, sowing all of your work. Sure looks like 8. (4 + ) 2 16 lim ( ) (8 + ) = Consider a iker wose position along a trail is given by 0 miles at time t = 0 minutes, 3 mi at time t = 1 our, 7 mi at time t = 2 ours, and 8 mi at time t = 3 ours. Calculate te

6 iker s average speed over te first our of te ike. Ten estimate te iker s speed at time t = 1 our. Average speed over first our = (3 0)/(1 0) = 3 mp. To estimate speed at time t = 1, we ll calculate anoter difference quotient on te oter side and ten take te average. Average speed over second our = (7 3)/(2 1) = 4 mp. We average tese to get (4 + 3)/2 = approximately 3.5 mp was te iker s speed at time t = Te dept D (in inces) of water in a swimming pool being filled is given by D = f(t) were t is in minutes since te ose was turned on. (a) Is f (t) positive or negative? Wat are te units of f (t)? f (t) is positive since te dept of water is increasing. Te units of f (t) are inces/minute. (b) Wat would it mean to say tat f(20) = 5 and f (20) = 0.1? f(20) = 5 means tat twenty minutes after te ose was turned on, te pool is full to a dept of 5 inces. f (20) = 0.1 means tat after te ose as been on for 20 minutes, te level of water is rising at about 0.1 inces/minute. (c) Wat are te units of te number 0.1 above? inces/minute 24. A study of 100 Boulder residents concluded tat te number N of appy residents on any given day was given by N = f(), a function of te number of ours of sunsine () during tat day. (a) Suppose we found tat f(3) = 40. Wat are te units of 3? Wat are te units of 40? Te units of 3 are ours of sunsine. Te units of 40 are number of appy people (out of 100). (b) Suppose tat f (8) = 1. Wat are te units of 8? Wat are te units of 1? Te units of 8 are ours of sunsine. Te units of 1 are additional appy people per additional our of sunsine. (c) Suppose tat f 1 (50) = 4.5. Wat are te units of 50? Wat are te units of 4.5? Te units of 50 are appy people (again out of 100). Te units of 4.5 are ours of sunsine. 25. Consider te function f(x) = 20 2x + 4. (a) Use te limit definition of te derivative to find f (6).

7 f (6) f(6 + ) f(6) 20 2(6 + ) (6) (16 + 2) 20(16) 40 = for small values of (b) At wat x-values, if any, is tis function not differentiable, and wy? Tis function is not differentiable at x = 2 because its grap as a corner tere. 26. Consider te function f(x) = { x wen x 0 x wen x < 0 At wat x-values, if any, is tis function not continuous? At wat x-values, if any, is tis function not differentiable, and wy? Tis function is continuous everywere. It is not differentiable at x = 0 since te derivative of 1 x is 2 wic goes to as x 0. Te grap as a vertical tangent line at x = 0. x 27. Derivative practice: in eac problem, find te equation of te derivative function. (a) f(x) = x 19 f (x) = 19x 18 (b) f(x) = x f (x) = 2x 5 (c) f(x) = π 3 (constant!) f (x) = 0 (d) f(x) = x + 1 x f (x) = 1 2 x 1 x 2 (e) f(x) = 1 + x + x2 + x 3 + x 4 x 3 f (x) = 3 x x x (f) f(x) = 1 x 2 f (x) = 2 x 3 (g) f(x) = (x + 1)(2x 1) = 2x 2 + x 1 f (x) = 4x + 1 () f(x) = kx n f (x) = knx n 1 (i) f(x) = 5 x x 5 f (x) = 5 x ln(5) 5x 4 (j) f(x) = x x 3 f (x) = 96x x 4

8 28. If te position of a particle at time t is given by te formula s(t) = t 3 t, wat is te velocity of te particle at time t = 1? v(t) = s (t) = 3t 2 1 v(1) = 3 1 = 2 (m/s?) 29. A rock falling from te top of a vertical cliff drops a distance of s(t) = 16t 2 feet in t seconds. Wat is its speed at time t? Wat is its speed wen it as fallen 64 feet? v(t) = s (t) = 32t Wen it as fallen 64 feet, 64 = 16t 2 so t = 2. v(2) = 64 ft/s.

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