QUIZ ON CHAPTERS 1 AND 2 - SOLUTIONS REVIEW / LIMITS AND CONTINUITY; MATH 150 SPRING 2017 KUNIYUKI 105 POINTS TOTAL, BUT 100 POINTS = 100%

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1 QUIZ ON CHAPTERS AND 2 - SOLUTIONS REVIEW / LIMITS AND CONTINUITY; MATH 50 SPRING 207 KUNIYUKI 05 POINTS TOTAL, BUT 00 POINTS = 00% ) For a), b), and c) below, bo in the correct answer. (6 points total; 2 pts. each) a) Let f ( ) = 4 + cos( ). The function f is even odd neither Dom f b) Let g Dom g c) Let h =., f ( ) = ( ) 4 + cos( ) = 4 + cos( ) = f ( ). = + sin( ). The function g is even odd neither =., g( ) = + sin = sin = g( ). = 5 +. The function h is even odd neither. A countereample to h being even or odd: h 2 and h( 2) are neither equal nor opposite. h 2 =, but h( 2) = 29. 2) Fill in the blanks. Find rules for functions f and g so that ( f g) ( ) = f ( g( ) ) =. (We are decomposing a composite function.) 2 + g( ) = 2 +, f ( u) = u (Do not let f or g be the identity function.) (2 points) There are different possible answers, but the above would be a reasonable choice. ) Complete the Identities. Fill out the table below so that, for each row, the left side is equivalent to the right side, based on the type of identity (ID) given in the last column. (8 points total; 2 points each) Left Side Right Side Type of Identity (ID) + tan 2 u cos u sin u ( u) cos u + v sin u v cos 2 sec 2 ( u) Pythagorean ID cos( v) sin( u) sin( v) Sum ID cos( v) cos( u) sin( v) Difference ID + cos( 2u) 2 or 2 + cos( 2u) 2 Power-Reducing ID (PRI) 4) Write any two of the three different versions of the Double-Angle Identity (ID) for cos 2u that were listed in Chapter. (4 points) Any two of these three: = cos 2 ( u) sin 2 u cos 2u The Pythagorean Identity sin 2 ; cos 2u = 2sin 2 ( u); cos( 2u) = 2cos 2 ( u) ( u) + cos 2 ( u) = is used to get from the first to the others.

2 tan sin 2 5) Verify the identity sin( 2) tan( ) Q.E.D. = 2sin( )cos sin( cos = ( 2cos2 ) using the Chapter identities. (5 points) Double-Angle ID Quotient ID = 2 sin ( ) cos cos sin = 2cos 2 ( ) 6) Fill out the table below. Use interval form (the form using parentheses and/or brackets) ecept where indicated. You do not have to show work. (6 points) Domain Range f cos (, ), tan( ) csc( ) Use set-builder form. π 2 + πn ( n ) Use set-builder form. { πn ( n ) } 7) In parts a) through i), consider the function f, where f and the graph of y = f = in the usual y-plane. (28 points total) a) Find f 2. Show all rigorous work, as in class. (7 points) f = 2 2 b) Find f (5 points) = 4 7 Limit Form (, ) (,, ) , ( ) ( + 2) ( 2) ( +) 2. Show all rigorous work, as in class. Use the Division Method to find a long-run it as. Divide the numerator and the denominator by 2 (or by for the simplified form on the right), the highest power of in the denominator.

3 (cont.) f 2 4 = or ( from Part a) ) ( This is the ratio of the given leading coefficients; see f). ) c) Which of the following is true? Bo in one: (2 points) i. f is continuous at = 2. ii. f has a removable discontinuity at = 2. iii. f has a jump discontinuity at = 2. iv. f has an infinite discontinuity at = 2. f is a rational function, and 2 Dom( f ), so f is continuous at 2. d) Which of the following is true? Bo in one: (2 points) i. f is continuous at =. ii. f has a removable discontinuity at =. iii. f has a jump discontinuity at =. iv. f has an infinite discontinuity at =. f is rational, and = The corresponding problem factor, + yields the it form nonzero real number 0 (divided) out in the denominator when the original f + f , is not completely canceled epression is simplified. Limit Form 5/ = f Limit Form 5/ =. 0

4 e) Which of the following is true? Bo in one: (2 points) i. f is continuous at = 2. ii. f has a removable discontinuity at = 2. iii. f has a jump discontinuity at = 2. iv. f has an infinite discontinuity at = 2. In Part a), we determined that 2 f However, f 2 denominator of eists. is undefined ( 2 Dom( f )) because Therefore, f has a removable discontinuity at = 2. f) What is the horizontal asymptote of the graph of y = f equation. Answer only on f) is fine. ( points) is a factor of the? Write its f is a rational function, so its graph has at most one horizontal asymptote (HA). From Part b), f =, so the sole HA has equation: y =. Note: For both original and simplified epressions for f ( ), the numerator and denominator have equal degrees, and the ratio of their leading coefficients is. g) Where is the hole on the graph of y = f ( )? Use, y coordinates. Answer only on g) is fine. ( points) form to write its From Part a), f 2 = 4. f is discontinuous at = 2 (it is undefined there), 7 yet 2 f eists, so there is a hole at: 2, 4 7 Observe that the ( 2) factor is completely canceled (divided) out in the denominator when the original f ( ) epression is simplified. h) What is the -intercept of the graph of y = f ( )? (2 points) Set y = 0. The real zeros of f are those of = , or + 2 ( + 2 ): + 2 = 0 ( 2) = 2 The -intercept is at ( 2, 0). i) What is the y-intercept of the graph of y = f ( )? (2 points) Set = 0 y = f ( 0) = The y-intercept is at ( 0, 2) = = 2 = 2 = 0 0

5 The graph of y = f ( ) is below. Note the horizontal asymptote (HA) at y =, the vertical asymptote (VA) at =, and the hole at 2, 4 7. There is continuity at = 2 ; in fact, the -intercept is at ( 2, 0). The y-intercept is at ( 0, 2). = ) Let f graph of y = f. What is the equation of the horizontal asymptote of the in the usual y-plane? ( points) The sole horizontal asymptote (HA) is given by y = 0, because = Alternatively, = 0. Note that is proper and rational, and we are taking its long-run its. (By proper, we mean that the numerator s degree, 4, is less than the denominator s degree, 0. Think: Bottom-heavy. ) 9) Find the following its. Bo in your final answers. (22 points total) a) r r r r r 2. Show all work, as in class. (0 points) r We have a 0 0 r 2 r it form. We will rationalize the numerator. ( r 2 ) ( r 2 + ) r r 2 + r ( r 2) 2 2 ( r) r 2 + Observe: r r ( r 2) 9 ( r) r 2 + ( r 2 is defined near r =. ) r and ( r) are opposites. Their quotient is. r 2 + = 2 + = 9 + = + = 6 ( ) r ( r) r 2 + ( )

6 b) cos ( ). Show all work, as in class. (6 points) cos( ) = 0. Prove this using the Sandwich / Squeeze Theorem: cos( ) ( ) More precisely: Observe that As, = 0 and > 0, > 0. cos So, by Sandwich/ Squeeze Thm. Divide all three parts by ( > 0) = 0, so by the Squeeze Theorem, 6 c). Show all work, as in class. (6 points) = 6 + ( ) ( + 2) + 5 cos Limit Form ( ) = 0. 0) (2 points). True or False: It is possible that for a function f, a f but f a = eists does not (meaning f ( a) is undefined). Bo in one: True False For eample, see Section 2., Eample 7; here, let s use the letter f instead of g. Let f ( ) = + ( ). f = 6 (the it eists), yet f is undefined. = L a, L ) Write a precise ε-δ definition of f a. Assume f is defined on a punctured neighborhood of a. (7 points) For every / any / all ε > 0, δ > 0 such that if 0 < a < δ, then f () L < ε. there eists Alternatively: then f ( ) is in ( L ε, L + ε)

7 2) Let f ( ) = 4, 2 2, > 2. (6 points total). (2 points) a) Evaluate 2 + f Since we are letting approach 2 from the right, the case > 2 is the only relevant case here, and the rule f ( ) = b) Evaluate f ( ). (2 points) 2 is the only relevant rule here. f ( 2 ) 2 + = 2 2 = Since we are letting approach 2 from the left, the case 2 is the only relevant case here, and the rule f = 4 is the only relevant rule here. 2 f = ( 2) 4 = c) Which of the following is true? Bo in one: (2 points) ) Let g( t) = i. f is continuous at = 2. ii. f has a removable discontinuity at = 2. iii. f has a jump discontinuity at = 2. iv. f has an infinite discontinuity at = 2. The right-hand it from a) and the left-hand it from b) as 2 both eist, but they are not equal. Therefore, f has a jump discontinuity at = 2. t 4 + t 7. What is the domain of g? Write your answer in interval form (the form using parentheses and/or brackets). Note: g is continuous on the domain interval(s). (4 points) The denominator, t 4, is real and nonzero t 4 > 0 t > 4. t 7 is real everywhere on. Dom( g) = ( 4, ) 4) (2 points). True or False: If f is a polynomial function on such that f = 0 and f ( 7) = 20, then the equation f ( ) = 5 has a real solution. Bo in one: True False f is polynomial on and hence continuous on, in particular on the interval, , 20. By the Intermediate Value Theorem (IVT), f in, 7. = 5 has a real solution