Math 102 TEST CHAPTERS 3 & 4 Solutions & Comments Fall 2006

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1 Mat 102 TEST CHAPTERS 3 & 4 Solutions & Comments Fall 2006 f(x+) f(x) For f(x) = x 2 + 2x 5, find ))))))))) and simplify completely. NOTE: **f(x+) is NOT f(x)+! f(x+) f(x) (x+) 2 + 2(x+) 5 ( x 2 + 2x 5 ) f(x+) is NOT f(x) (x+)! )))))))))) = ))))))))))))))))))))))))))))))))) x 2 + 2x x ( x 2 + 2x 5 ) = )))))))))))))))))))))))))))))))))))))) x 2 + 2x x ( x 2 + 2x 5 ) Many terms cancel. Notice 5 ( 5) = 0 = )))))))))))))))))))))))))))))))))))))) 2x = )))))))))))))))))))))))))))))))))) (2x + + 2) A factor must be a factor of te entire = )))))))))))))))))))) numerator and of te entire denominator in order to reduce (in order to cancel ). = 2x Use te grap of te function at rigt to answer te following questions. a. Wat is f( 2)? f ( 2) 2 (See on grap.) 6 b. Wat is te domain of f? [ 6,4.5] (estimated) c. Wat is te range of f? [ 4, 3] d. On wat interval(s) is f increasing? f increases wen 6 x 3 & 1.2 x 3 (rougly) On te interval [ 6, 3] & on [~1.2, ~3] e. Wat is te average rate of cange of f on te interval [ 3,4]? Te eigt of te grap at x= 3 is y=3. Te eigt at x=4 is y= 1. (See So wile x increased 7 units, y dropped from 3 to 1 (3 1 = 4). Te slope of te line connecting te two points is tus: y y 2 y ))) = ))))) = ))))) = )) x x 2 x 1 4 ( 3) points on grap.)

2 Sketc te grap of y = 2(x+1) using transformations of a familiar function. 8 y = x 3 y = (x+1) 3 y = x 3 y = (x+1) 3 y = 2 (x+1) 3 1 Adding 1 to te argument sifts te grap 1 unit left. Multiplying by -1 flips te curve about te x-axis. Multiplying by 2 stretces it vertically Sketc te grap of f(x) = 2x 2 + 6x + 3. Label te coordinates of te vertex & y-intercept. Does tis function ave a maximum or minimum value? Wat is it? f(x) = 2 ( x 2 + 3x ) + 3 f(x) = 2 ( x 2 + 3x ) = 2 ( x +? ) = 2 (x ) Tis is clearly te squaring function... Sifted left 3 2, stretced by te factor 2, sifted down 3 2. Tus 3 2 is te minimum value. = 2 (x ) ANOTHER WAY to reac tis conclusion : An alternative is to know tat (0, 3) f(x) = ax 2 + bx + c as its extreme point at -b/2a, wic in tis case is -6/4 = Te leading coefficient in f(x) = 2x 2 + 6x + 3, wic is 2, informs us te parabola opens upward, so we know te extreme point is a minimum. Finally we evaluate f(x) at x = 3 2 : f( 3 2 ) = 2( 3 2 ) 2 + 6( 3 2 ) + 3 = to obtain te minimum value. ( 3 2, 3 2 )

3 10 5. Sketc te grap of te function f(x) = x 2 (x 2)(x + 1). Label all intercepts wit teir coordinates, and describe te end beavior of f. Tat f(x) is a 4 t -degree POLYNOMIAL* function is clear witout computing: f(x) = x 4 x 3 2x 2 f(x) = x 2 (x 2)(x + 1) Te ROOTS of f are 0,0, 2 and -1. f(x) and wen x is large (+ or ), f(x) is dominated by, and beaves like x 4, wit y rising towards infinity * Terefore, te grap of f is smoot, continuous, and beaves like its largest power as x 2x 2x Sketc te grap of f( x) = ))))) = )))) ))) x 2 1 (x+1)(x 1) Domain: all reals except 1 & 1 RATIONAL function wit vertical asymptotes at x=1 and x= 1. f(0) = 0... Tus (0,0) is bot x-intercept and y-intercept. Tere are no more intercepts. As x, denominator (degree 2) grows muc faster tan numerator (degree 1); function beaves like 2 x, wic we know 0 as x. So y=0 is orizontal asymptote. It elps a lot to determine weter f(x) is positive or negative on eac interval between points of interest. f(x)! + 0! + One final comment on tis: F is an ODD function; note te symmetry. x= 1 x=1 y=0 (0,0)

4 10 7. Solve te inequality, and express your answer in INTERVAL notation. 5 3 Q:! +! 0 + ))))) ))))) ))))))))))))))))))))))))))))))) x + 3 x ))))) ))))) 0 Q is + for all x between 3 & 1 and above 7. x + 3 x 1 Noting tat te fractions do not exist at 3 & 1 5 (x 1) 3 (x +3) eliminates 3 & 1 as solutions. )))))))) )))))))) 0 ( x+3)(x 1) (x 1)(x +3) Testing 7, we get 0 0, so 7 is a solution. 2x 14 )))))))))) 0 Te solution set for tis inequality is: ( x + 3) ( x 1) ( 3, 1) [7, ) critical values (were te signs of te factors can cange) are 7, 3, For P(x) = 2x 3 + 5x 2 15x + 6, a. List ALL te possible rational zeroes of P(x). P(x) as all integer coefficients, so any rational roots must be of te form p q, were p divides 6 and q divides 2. Te possibilities consist of all viable combinations of ± 6, 3, 2, 1...wic gives us te LIST: ± 1, 2, 3, 6, 1 2, 3 2 (Notice tere are twelve!) 2, 1 b. Use syntetic division to sow tat ½ is a zero of P(x). ½ Te zero remainder tells us x ½ is a FACTOR of P(x) Tus ½ is a ROOT (aka ZERO ) of P(x). c. Find ALL te zeroes of P(x). Simplify your answers. To find te remaining zeroes of P(x), we use te quotient, Q(x) = 2x 2 + 6x 12. Aside from x = ½, only values of x tat 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 te useful quadratic formula. x = 3 ± 9 4 ( 6 ) 3 ± 3 3 ))))))))))) = )))))))) 2 2 Te tird zero is te one we verified in part b: ½

5 10 9. Find a fourt-degree polynomial P(x)= a x 4 + bx 3 +cx 2 +dx + e wit REAL coefficients, and wit 0 a root (zero) of multiplicity 2, and wit 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 ^of multiplicity k So P(x) = (x 0) (x 0) (x (1+i)) (x (1 i)) = x 2 ( x 2 2x + 2) = x 4 2x 3 + 2x 2 ** See Details below. (Note: Te given specifications do not fully determine P. Any polynomial of te form P(x) = A ( x 4 2x 3 + 2x 2 ) would answer te request.) **Details: (x A) (x B) = x 2 + ( A B)x + AB...for any numbers A & B Here, A = 1 + i and B = 1 i, so A B = 1 i 1 + i = 2 and AB = ( 1+ i )( 1 i ) = 1 + i i i 2 = A rectangular storage space is to be enclosed wit 200 yards of fencing. One side of te storage space faces an existing fenced yard and does not require fencing. Let x be te lengt of te side tat projects out from te existing fance. a. Express te area of te storage space in terms of x. AREA = lengt widt x A(x) = (200 yd 2x) x = 2(100 -x) x 200yd 2x EG if x = 60 yd, ten 120 yd are used up by x-fences, so y can be only yd. If x = a yd, ten 2a yd are used up by x-fences, so only 200 2a remain for y. x b. For wat value of x will te area be maximum? We see te function above is a parabola, opening down, y = A(x) wit x-intercepts at 0 and 100. Terefore, te maximum value for A MUST occur wen x is midway between tese zeroes.... wen x = 50 (yd) Alternatively, we can observe A(x) = 2x x and complete te square or use te formula to obtain x = 200/ (Not requested, but te maximum area is 50 yd 100yd = 5000 square yards.)