Pre-Calculus. Test Solutions

Math 102 10 TEST CHAPTERS 3 & 4 Solutions & Comments f(x+h) – f(x) = h )))))))))) = = find x2 + 2xh + h2 + 2x + 2h – 5 — ( x2 + 2x – 5 ) h )))))))))))))))))))))))))))))))))))))) x2 + 2xh + h2 + 2x + 2h – 5 — ( x2 + 2x – 5 ) Many terms “cancel”. )))))))))))))))))))))))))))))))))))))) h Notice – 5 — ( – 5) = 0 2xh + h2 + 2h h = )))))))))))))))))))))))))))))))))) = )))))))))))))))))))) = 10 f(x+h) – f(x) ))))))))) and simplify completely. h NOTE: **f(x+h) is NOT f(x)+h ! (x+h)2 + 2(x+h) – 5 — ( x2 + 2x – 5 ) f(x+h) is NOT f(x) (x+h)! ))))))))))))))))))))))))))))))))) h 1. For f(x) = x2 + 2x – 5, h (2x + h + h 2) A factor must be a factor of the entire numerator and of the entire denominator in order to reduce (in order to “cancel”). 2x + h + 2 2. Use the graph of the function at right to answer the following questions. 6 a. What is f(–2)? f (–2) . 2 (See b. What is the domain of f? [–6,4.5] (estimated) c. What is the range of f? [–4, 3] d. On what interval(s) is f increasing? on graph.) -6 5 f increases when –6# x# –3 & 1.2# x# 3 (roughly) On the interval [–6,–3] & on [~1.2, ~3] e. –5 What is the average rate of change of f on the interval [–3,4]? The height of the graph at x= –3 is y=3. The height at x=4 is y= –1. (See So while x increased 7 units, y dropped from 3 to –1 (3 – –1 = 4). The slope of the line connecting the two points is thus: –1 – 3 –4 ªy y2 – y1 ))) = ))))) = ))))) = )) ªx x2 – x1 4 – (–3) 7 points on graph.) 3 10 3. Sketch the graph of y = – 2(x+1) 8 y = x3 using transformations of a familiar function. y = (x+1)3 y = x3 y = (x+1)3 y =–2 (x+1)3 1 Adding 1 to the argument shifts the graph 1 unit left. 10 Multiplying by -1 “flips” the curve about the x-axis. Multiplying by 2 stretches it vertically. 4. Sketch the graph of f(x) = 2x2 + 6x + 3 . Label the coordinates of the vertex & y-intercept. Does this function have a maximum or minimum value? What is it? f(x) = 2 ( x2 + 3x = 2 ( x + ? )2 ) +3 f(x) +3 = 2 ( x2 + 3x + 9'4 ) +3 – 9'2 = 2 (x + 3'2 )2 +3 – 9'2 = 2 (x + 3'2 )2 – 9'2 This is clearly the squaring function... Shifted left 3'2, stretched by the factor 2, shifted down 3'2 . Thus –3'2 is the minimum value. ANOTHER WAY to reach this conclusion : An alternative is to know that (0, 3) f(x) = ax2 + bx + c has its extreme point at -b/2a, which in this case is -6/4 = - 3'2 . The leading coefficient in f(x) = 2x2 + 6x + 3 , which is [ 2, informs us the parabola opens upward , so we know the extreme point is a minimum. Finally we evaluate f(x) at x = 3'2 : 3 2 3 3 f(3'2) = 2( '2) + 6( '2) + 3 = – '2 ... to obtain the minimum value. ( –3'2 , –3'2 ) 10 5. Sketch the graph of the function f(x) = x 2 (x–2)(x + 1). Label all intercepts with their coordinates, and describe the “end behavior” of f. That f(x) is a 4th-degree POLYNOMIAL* function is clear without computing: f(x) = x 4 – x3 – 2x2 f(x) = x2 (x–2)(x + 1) The ROOTS of f are 0,0, 2 and -1 . f(x) + 0 –1 – 0 – 0 0 + 2 v v º and when x is large (+ or – ), f(x) is dominated by, and behaves like x4, with y rising towards infinity. -1 0 2 * Therefore, the graph of f is smooth, continuous, and behaves like its largest power as x 64 10 6. Sketch the graph of f( x) = 2x = )))) 2x ))) x – 1 (x+1)(x–1) ))))) 2 Domain: all reals except –1 & 1 RATIONAL function with vertical asymptotes at x=1 and x=–1. f(0) = 0 .... Thus (0,0) is both x-intercept and y-intercept. There are no more intercepts. As x64, denominator (degree 2) grows much faster than numerator (degree 1); function behaves like 2'x , which we know 6 0 as x64. So y=0 is horizontal asymptote. It helps a lot to determine whether f(x) is positive or negative on each interval between points of interest. f(x) – ! + 0 – ! + One final comment on this: –1 0 1 F is an ODD function; note the symmetry. x= –1 y=0 x=1 (0,0) 10 7. Solve the inequality, and express your answer in INTERVAL notation. 5 3 $ ))))) x+3 x–1 Q: ))))) 5 — x+3 3 $ 0 x–1 ))))) — ! –3 + ! — 0 7 + ))))))))))))))))))))))))))))))) 1 Q is + for all x between –3 & 1 and above 7. ))))) 5 (x – 1) 3 (x +3) )))))))) – )))))))) $ 0 ( x+3)(x –1) (x –1)(x +3) Noting that the fractions do not exist at –3 & 1 eliminates –3 & 1 as solutions. Testing 7 , we get 0 $ 0, so 7 is a solution. 2x – 14 $ 0 ( x + 3) ( x – 1) The solution set for this inequality is: )))))))))) úú (–3, 1) c [7, 4 ) critical values (where the signs of the factors can change) are 7, –3, 1 . 10 8. For P(x) = 2x3 + 5x 2 – 15x + 6, a. List ALL the possible rational zeroes of P(x). P(x) has all integer coefficients, so any rational roots must be of the form p 'q , where p divides 6 and q divides 2. The possibilities consist of all viable combinations of ± 6, 3, 2, 1 ...which gives us the LIST: ± 1, 2, 3, 6, 1'2 , 3'2 (Notice there are twelve!) 2, 1 b. ½ Use synthetic division to show that ½ is a zero of P(x). 2 2 c. 5 1 6 –15 3 –12 6 –6 0 The zero remainder tells us x – ½ is a FACTOR of P(x). Thus ½ is a ROOT (aka “ZERO”) of P(x). Find ALL the zeroes of P(x). Simplify your answers. To find the remaining zeroes of P(x), we use the quotient, Q(x) = 2x 2 + 6x – 12. Aside from x = ½, only values of x that make Q(x) = 0 will make P(x) = 0. 2x 2 + 6x – 12 = 0 x 2 + 3x – 6 = 0 ...Does not factor. So we resort to the useful quadratic formula. x = –3 ± q%%9%%!%%4%(!%6%%)% ))))))))))) 2 = –3 ± q%3%%3% )))))))) 2 The third zero is the one we verified in part b: ½ 10 9. Find a fourth-degree polynomial P(x)= a x4 + bx3 +cx2 +dx + e with REAL coefficients, and with 0 a root (zero) of multiplicity 2, and with 1+ i a zero or root. \ \ \ (x – 0) is a factor \ \ ... twice, and (x – (1+i)) is a factor and (x – (1– i)) is a factor. Since any number r is a ROOT of P if and only if (x – r) is a FACTOR of P. ^of multiplicity k So P(x) = (x –0) (x–0) (x – (1+i)) (x – (1– i)) = x2 ( x2 – 2x + 2) ^of multiplicity k ** See Details below. = x4 – 2x3 + 2x2 (Note: The given specifications do not fully determine P. Any polynomial of the form P(x) = A ( x4 – 2x3 + 2x2) would answer the request.) **Details: (x – A) (x – B) = Here, A = 1 + i and B = 1 – i, so – A – B = – 1 – i –1 + i = –2 10 x2 + (– A – B)x + AB ...for any numbers A & B and AB = ( 1+ i )( 1 – i ) = 1 + i – i – i2 = 1 +1 10. A rectangular storage space is to be enclosed with 200 yards of fencing. One side of the storage space faces an existing fenced yard and does not require fencing. Let x be the length of the side that projects out from the existing fance. a. Express the area of the storage space in terms of x. AREA = lengthCwidth A(x) = (200 yd – 2x) x = 2(100 -x) x x 200yd – 2x EG if x = 60 yd, then 120 yd are used up by x-fences, so “y” can be only 200 ! 120 yd. If x = a yd, then 2a yd are used up by x-fenc es, so only 200 – 2a remain for “y”. b. x For what value of x will the area be maximum? We see the function above is a parabola, opening down, with x-intercepts at 0 and 100. Therefore, the maximum value for A MUST occur when x is midway between these zeroes. ... when x = 50 (yd) y = A(x) 0 Alternatively, we can observe A(x) = –2x2 + 200x and complete the square or use the formula to obtain x = –200/4. (Not requested, but the maximum area is 50 yd C100yd = 5000 square yards.) 100