Math 102 Practice Final Solutions

Math 102 Practice Final 1 Practice Math 102 Common Final Exam 1. For A = {x|x < 1} and B = {x|x < 5}, find A1 B . Express your answer in interval notation. Set A is the real numbers less than 1: Set B contains all numbers up to 5: <))))))))))))B))))))))))))))))> <))))))))))))))))))))))))C))))))> A 1 B = { x | x is both < 1 and < 5 } = {x| x < 1} = A = (–4 ,1) . A 1 B = A because A is a subset of B (everything in A is included in B). 2. Simplify completely. Express your answer without negative exponents. (2x4y–4/5)5(16y–4)–1/4 First we give ou rselve s a little sp ace to thin k in, then d istribute th e expo nents, 5 and –1/4: (2 x4 y–4/5 )5 ( 16 y –4 )–1/4 = 25 x20 y–4 ( 16 y –4 )–1/4 = 25 x20 y–4 16–1/4 y –4 (–1/4) = 25 x20 y–4 16–1/4 y = –1/4 5 4 –1/4 2 16 20 x Regro up to facil itate com bining li ke factor s... –4 y y 20 x y–4 y = 2 (2 ) = 25 2–1 x20 y–4 y = 24 x20 y–3 24 x20 )))))))) y3 = 3. 5 e.g. (y–4/5 )5 = y(–4/5)5 = y–4 Factor completely. This s implific ation req uires on ly one o r two step s, ... which were drawn ou t to reveal all details x3/2 + 6x1/2 + 9x–1/2 x1/2 (x3/2 + 6x1/2 + 9x–1/2 ) )))) x1/2 Observation: Multiplication by x 1/2 would make that a lot nicer! (x3/2+1/2 + 6x1/2+1/2 + 9x–1/2+1/2 ) x ))))))))))))))))))))))))))))))))) 1/2 (x2 + 6x1 + 9 ) x )))))))))))))))) 1/2 (x + 3) 2 x ))))))))) 1/2 4. Divide as indicated. Express your answer in the form a + bi. 6i + 9 )))))) 1 – 3i = (6i + 9)(1 + 3i) 9C1 — 6C3 + 9C3i +6i ))))))))))))) )) = )))))))))))))))))))) (1 – 3i)(1 + 3i) 1+9 = –.9 +3.3i Math 102 Practice Final 5. 2 Solve. Express your answer in interval notation. x < 2 x+4 )))))) Resist the temptation to multiply both sides. compare to 0, not 2 !! and SIMPLIFY x x+4 – 2 < 0 x x+4 2(x+4) – )))))) (x+4) )))))) )))))) x – 2(x+4) x+4 ))))))))))) < 0 –8 –x – 8 x+4 < 0 quotient –x – 8 x+4 )))))) 6. < 0 –4 )))))) |)))))))) |))))) + – – – – + – + – check endpoints: at – 8: 0 < 0 (inequality is true) at – 4: inequality false Statement is true for x in (–4 ,–8) U [– 4,4 ) Solve. Express your answer using interval notation. 2| (½)x + 3| + 3 < 15 I like to see the x with coefficient 1, so I go ahead and multiply 2| (½)x + 3| | x + 6 | + 3 < 15 |x + 6 | < 12 Here I LOOK at it and solve: |x – –6 | < 12 ... says “distance between x & –6" < 12 [ –6 – 12, –6 + 12] [ –18 , 6 ] But this works, too: |x + 6 | < 12 if and only if –12 < x + 6 –18 < 7. x < 12 < 6 so x is in [ –18 , 6 ] Let A = (2,–5) and B = (–1,3) be points in the xy-plane. a. Find the length of segment AB. (simplify the answer) Length of AB = ))))))))))) / ()x)2 + ()y)2 = ( 3 2 + 8 2 )1/2 = 731/2 b. Find the midpoint of segment AB. Midpoint of AB = (halfway between 2 & –1, halfway between –5 & 3) =( ' 2 , ( –5 ( 2 + –1 ) = ( ½ , –1 ) + 3) '2 ) Math 102 Practice Final 8. 3 Find the center and radius of the circle with equation ... This is most easily done by completing the square. x 2 + y 2 – 4x – 6y – 3 = 0 x 2 – 4x + y 2 – 6y ( x – what) 2 + ( y – what) 2 x 2 – 4x + 4 ( x – 2) 2 = + Y2 = Preparing to complete the sq uare 3 + whatever I need.... + y 2 – 6y + 9 + ( y – 3) 2 X2 = 3 = 3+4+9 3 + 13 = 16 Fill in the gaps... and compa re to the essen tial equation of a c ircle So we see this is the equation of a circle with center at (2,3) and radius 4. 9. The manager of a cheap plastic furniture factory finds that it costs $2400 to produce 200 chairs in a day, and $6200 to produce 400 chairs in a day. Assuming that the relationship between cost (y) and the number of chairs produced (x) is linear, find an equation that relates x and y. Reduced to its essence, this problem asks us to find the equation of a line through (x,y) = (200,2400) and (400,6200). SLOPE = )y/)x = 3800/200 = 19 Here I use the point-slope form: the equation is y – y0 = 19 (x – x0) where (x 0,y 0) is any point on the line y – 2400 = 19 (x – 200) y = 19x – 1400 Here I use the slope-intercept form: to find b, I take advantage of the fact that when x = 200, y = 2400: 10. THIS IS SUFFICIENT But you may also write this. y = 19x + b 2400 = 19(200) + b So b = 2400 - 19(200) = -1400. For f(x) = 2x2 – 5x + 3, find & simplify completely: f(x+h) – f(x) ))))))))))) = h 2(x+h)2 – 5(x+h) + 3 — ( 2x2 – 5x + 3) h )))))))))))))))))))))))))))))))))))))) 2( x2 + 2xh + h2 ) – 5(x+h) + 3 — ( 2x2 – 5x + 3) h = ))))))))))))))))))))))))))))))))))))))))))))))) = )))))))))))))))))))))))))))))))))))))))))))))) = = Distribute that — !! 2x2 + 4xh + 2h2 – 5x – 5h + 3 — 2x2 + 5x — 3 h 4xh + 2h2 – 5h ))))))))))))))))))))) h 4x + 2h – 5 ... And there you have it, in excruciating detail.... Math 102 Practice Final 11. Sketch the graph of f(x) = r%x%% Use transformations to sketch the graph of y = 4 –x r%%%%% –2 I used intermediate y = r%%–%x%% x 12. Find the formula for f–1(x) given f(x) below: 1 f(x) = ))))))) 3x – 2 1 y = )))))) 3x – 2 We solve for x in terms of y. this will reveal the function that, given y, finds x. 3xy – 2y = 1 3xy = 1 + 2y 1 + 2y x = ))))))) 3y 13. We have found x in terms of y. The connection to the inverse is... f –1(y) = x... f–1(y) = ))))))) 1 + 2y 3y But it does not matter what the variable is called... f–1(x) = ))))))) 1 + 2x 3x It’s still the same function.... Sketch the graph of P(x) = –x3 + 4x2 – 4x . Label all intercepts and describe the end behavior. As x 64, P(x) 6 – 4 since as x grows larger, the leading term dominates; that is, as x64, – x3 6!4 faster than 4x2 – 4x 6 +4 , so the sum of the two 6!4. Similarly, as x 6 – 4 , P(x) 6 + 4 . . . . . . . . . . . . . . . . . . . . . . P(0) = 0, so that’s the y=intercept, at (0,0) . . . . . . . . . . . . . . . . . . x-intercepts: . . . . . . . . . . . . . . . . . . . . . . . . –x3 + 4x2 – 4x = 0 . . . . . . . . . 2 –x (x – 4x + 4) = 0 . . . . . . . . . . . . . . . . . . . . . . –x (x – 2)(x – 2) = 0 . . . . . . . . . . . . . . . . . . ... we see x-intercepts at 0, 2, and 2. . . . . . . . . . . . . . . . . . . . . . (make allowances for the graphing...) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Math 102 Practice Final 14. 5 For P(x) = 2x3 + 8x2 + 3x – 10 , a. List all the possible rational zeros of P(x) Because the coefficients of this polynomial are all integers, we know that any rational roots must be of the form p/q where p divides 10, and q divides 2. This yields: ±1, 2, 5, 10, 1/2, and 5/2 b. Use synthetic division to show that – 2 is a zero of P(x). –2 2 2 c. 8 –4 4 3 –8 –5 –10 10 0 This tells us P(x) = (x +2)(2x2 + 4x – 5) So –2 is a root of P. Find all the zeros of P(x). Simplify your answers. The remaining roots of P must be found in (2x2 + 4x – 5) Using the quadratic formula to solve 2x2 + 4x – 5 = 0... we find: x = –4 ± q %1% 6% %+ % %4 %0 % )))))))))) 4 = –1 ± 14 2 q %% % % ))) Of course the third root of this cubic polynomial is –2, the root we found above. 15. Find a fourth degree polynomial P(x) with 1+2i a zero, 1 a zero of multiplicity 2, and a constant term of 10. Express your answer in the form P(x) = ax3 + bx3 + cx2 + dx + e, where a,b,c,d,e, are integers. To be integers, a,b,c,d,e must all be real numbers. Since then P has only real coefficients, any complex roots must be paired with their conjugates as roots. Thus we know the four roots of P must be 1+2i, 1–2i, 1 and 1. And P must be the product of these factors: P(x) = A (x – (1+2 i) ) (x – (1–2i) ) (x – 1) (x –1) where A is some non-zero constant. This is: P(x) = A (x2 – 2x + 5) (x2 – 2x + 1) which has constant term 5A. Since they want the constant term to be 10, A had better be 2. P(x) = 2 (x2 – 2x + 5) (x2 – 2x + 1) P(x) = 2x4 – 4x3 + 20x2 – 24x + 10 Math 102 Practice Final 16. 6 Find the x and y intercepts. (express as ordered pairs) Find the vertical asymptote. (write the equation) Find the horizontal asymptote. (write the equation) Sketch the graph.... for: x– 6 y = )))))) x+ 2 Domain: y is defined for all values except x = –2 As x approaches –2, x– 6 ÿ –8, while x+2 ÿ 0, so the quotient ÿ ± 4 Thus there is a vertical asymptote at x = –2. y=1 As x ÿ4 y ÿ1 As x ÿ4 y ÿ1 Thus the function has a horizontal asymptote at y = 1 y-intercept: when x = 0 y = –3 x-intercept: when x = 6, y = 0 17. 18. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x= –2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (6,0) (0,–3) Sketch the graph of f(x) = – ex–2 (Label the asymptote and one point on the graph.) . . . . . . . . . . . . . . . . Working from the known graph of . . . . . . . . . . . . . . . . . . y = ex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . y = ex–2 is the same graph, shifted to . . . . . . . . . . . . . . . . the right two units . . . . . . . . . . . . . . . . . . y=0 . . . . . . . . . . . . . . . . . . . . . . . . . . (2,–1) . . . . . y = –ex–2 is the same graph, reflected through the x-axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simplify completely: 5 4 log4(2x) – log4 x + log4(4x) log4(2x)4 – log4 x5 + log4(4x) (2x)4 (4x) log4 ))))))))) x5 log4 64 . . . . . . Method A: Using the properties of logarithm functions, we can combine these terms into one log expression, and see if the result can be simplified. It can. Math 102 Practice Final 18. 7 Simplify completely: Method B: Using the properties of logarithm functions, we can expand & separate these terms so the log4x terms are alone. 4 log4(2x) – log4 x5 + log4(4x) 4 log42+ 4log4 x – 5log4x + log44 + log4x log416 + log44 + (4 – 5 + 1) log4x 2 + 1 + 0 19. Solve for x (Give an exact answer.): ex – 2 = 3x Std. Operating Procedure: exponential function, ln(e x – 2) = ln(3 x ) haul out the logs.... (x – 2) ln(e) = x ln (3) x–2 and use those logarithmic properties! = x ln 3 ... and keep in mind that ln3 is just a number. x – xln3 = 2 ( 1 – ln 3) x = 2 x = 2/(1– ln3) 20. 21. Well, that’s an exact answer. Solve for x: log 6 (x–3) + log 6(x+2) = 1 log 6 ((x–3)(x+2)) = 1 ((x–3)(x+2)) = 61 x2 – x – 6 = 6 x2 – x – 12 = 0 (x – 4)(x+3) = 0 x = 4 or –3 Better to have one log expression than two, Now we can “undo” the log 6 using exponential base 6. CHECK THOSE SOLUTIONS!!!! Only x = 4 is a valid solution. $2000 is invested in an account that pays 3% (annual percentage rate) compounded continuously. How long will it take for the balance to reach $3000? When amount P0 is invested at r % interest compounded N times per year, the value of the account at time t years is Nt P(t) = P0 ( 1 + r/N ) rt As N 64, this approaches P0 e ... the value achieved with continuous compounding. .03t So, the value of this account after t years is $2000 e and we want to solve for t so that this value will be $3000. .03t $2000 e e .03t .03 t t = $3000 = 1.5 = ln 1.5 = ln 1.5 years Ñ 13.52 years .03 )))))