MATH 121 FINAL Exam Review

Function Type Domain Rangeg(x) = 3x 2 − 20h(x) = 3 log 7 xk(x) =| 2x + 3 | −8j(x) = 4 (6 x )13. (Sections: Chapter 5 Toolbox, 5.1). Use properties of exponentsto show that y = 3 x − 4 is equivalent to y = 1 81 (3x ).14. (Sections: Chapter 5 Toolbox, 5.1). Use properties of exponentsto show that y = 6 −x ( ) 1 xis equivalent to y = .615. (Section: 5.1). At the end of an advertising campaign, the weeklysales (in dollars) declined, with weekly sales given by the equationy = 12, 000 ( 2 −0.08x) , where x is the number of weeks after the endof the campaign.(a) Determine the sales at the end of the campaign.(b) Determine the sales 6 weeks after the end of the campaign.(c) Does the model indicate that sales eventually reach $ 0? Explain.16. (Section 5.2). Between the years 1976 and 1998, the percent of momswho returned to the workforce within one year after they had a child isgiven by the equation, w(x) = 1.11 + 16.94 ln x, where x is the numberof years past 1970.(a) Find w −1 (x).(b) Use w −1 (x) to estimate the year in which 50 percent of momsreturned to the workforce within one year.17. (Section 5.3). Solve each equation algebraically. When necessary,round answers to four decimal places.(a) log 4 x = −2.

(b) 4 + log x = 10.(c) e (−2x+3) = 2(d) ln (−2x + 3) = 10(e) 2 ln x + 7 = ln (4x) + 1018. (Sections: 5.2, 5.3). Rewrite 2 log x + 5 log y − 8 log z as a singlelogarithm.( y 2 e 3x )19. (Sections: 5.2, 5.3). Rewrite ln as the sum, difference orz 3product of logarithms and simplify if possible.20. (Section: 5.5). Suppose $ 9,000 is invested for t years at 5.5% interestcompounded monthly.(a) Write an equation that gives the future value, S, of the investmentafter t years.(b) Find the future value of the investment in 4 years.(c) Find the number of years it will take the investment to double.21. (Sections: 6.1, 6.3). Let f(x) = 3x 3 + 18x 2 − 12x − 72. Use thisfunction to answer each question.(a) State the degree and leading coefficient of f(x).(b) Describe the end behavior of the graph of f(x).(c) Find all x such that f(x) = 0. Solve algebraically. Show work.(d) Use your calculator to create a complete graph of f(x). Sketchthe graph and list your window settings.(e) How many turning points does the graph of f(x) have?(f) How many inflection points does the graph of f(x) have?(g) At what point does a local maximum occur? round the coordinatesto one decimal place.22. (Section: 6.3). Solve algebraically.(a) 0.2x 3 − 20x = 0.(b) x 3 − 15x 2 + 56x = 0.

(c) 2x 4 − 3x 3 − 20x 2 = 023. (Sections: 7.1, 7.2). Solve the following system of equations algebraically.⎧⎪⎨⎪ ⎩2x − 3y + z = 23x + 2y − z = 6x − 4y + 2z = 224. (Section: 7.2). A car rental agency rent compact, midsize, and luxurycars. Its goal is to purchase 80 cars for a total of $1,822,000 and toearn a daily rental of $ 2,424 from all the cars. The compact cars cost$16,000 each earn $ 20 per day in rental, the midsize cars cost $ 22,000each earn $ 30 per day, and the luxury cars cost $ 38,000 each and earn$ 52 per day. Your task will be to find the number of each type of carthe agency should purchase to meet its goal.(a) Let x represent the number of compact cars purchased, y representthe number of midsize cars purchased and z represent the numberof luxury cars purchased. Write a system of equations to modelthis situation.(b) Write the augmented matrix for your system of equations.(c) Use your calculator to find the reduced-row echelon form of theaugmented matrix.(d) Find the number of each type of car the agency should purchaseto meet its goal.

MATH 121 Final Exam Review Solutions1. (a) (−23, −20)(b) Shift down 20 units, left 23 units and stretch by a factor of 5.(c) x = −25, x = −212. (a) x = −1 ± i√ 34(b) x = −3 ± i√ 63(c) x = ±2i(d) x = 4 ± 3i3. (a) x ≥ 0(b) yes; give explanation(c) C −1 (x) = −6 + √ x − 2064(d) 28 units4. For x < −2: decreasing, concave up with an open circle at (−2, 3);For −2 ≤ x < 2: line segment with closed circle at (−2, −4) and anopen circle at (2, 8);For x ≥ 2: increasing, concave down with a closed circle at (2, 2).5. One possibility is the graph of f(x) = −x 3 − 3.6. One possibility is the graph of g(x) = −x 2 + 2.17. Sketch the graph of h(x) =2−6; vertical asymptote at x = −2,(x + 2)horizontal asymptote at y = −6.8. (a) f[g(x)] = x(b) g[f(x)] = x(c) yes; give explanation9. w −1 (x) = 3 √x + 5210. (a) x = 4 and x = 7(b) x = −1 and x = 18(c) x = 20

11. (a) 350 units(b) $ 25,000(c) 150 ≤ x ≤ 55012. (a) g(x): quadratic, Domain: (−∞, ∞), Range: [−20, ∞)(b) h(x): logarithmic, Domain: (0, ∞), Range: (−∞, ∞)(c) k(x): absolute value, Domain: (−∞, ∞), Range: (−8, ∞)(d) j(x): exponential, Domain: (−∞, ∞), Range: (0, ∞)13. y = 3 x−4 = 3 x · 3 −4 = 3x3 4 = 3x81 = 181 (3x )( 1 x14. y = =6) 1x6 = 1 x 6 = x 6−x15. (a) $ 12,000(b) ≈ $8604(c) no; give explanation16. (a) w −1 (x) = e(x − 1.11)/16.94(b) 198817. (a) x = 1 16(b) x = 1, 000, 000(c) x ≈ 1.1534(d) x ≈ −11011.7329(e) x ≈ 80.3421( x 2 y 5 )18. logz 819. 2 ln y + 3x − 3 ln z20. (a) S = 9000 ( ) 12t1 + 0.05512(b) $ 11,209.06(c) ≈ 12.6 or 13 years

21. (a) n = 3; a = 3(b) concave down on left; concave up on right(c) x = 2, x = −2, and x = −6(d) possible window settings: [−10, 6, 2] x by [−100, 100, 25] y(e) 2 turning points(f) 1 inflection point(g) −4.3, 73.9)22. (a) x = 0 and x = 10 and x = −10(b) x = 0 and x = 7 and x = 8(c) x = 0 and x = −52 and x = 423. x = 2, y = 2, z = 4⎧⎪⎨ x + y + z = 8024. (a) 16x + 22y + 38z = 1822⎪⎩20x + 30y + 52z = 2424(b)(c)⎡⎢⎣⎡⎢⎣1 1 1 8016 22 38 182220 30 52 24241 0 0 350 1 0 280 0 1 17⎤⎥⎦(d) 35 compact, 28 midsize, and 17 luxury cars⎤⎥⎦