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THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MATH1020 General Mathematics (Fall 2016) Coursework 1 Suggested Solution Name: Student No.: Class: Final Result: I acknowledge that I am aware of University policy and regulations on honesty in academic work, and of the disciplinary guidelines and procedures applicable to breaches of such policy and regulations, as contained in the website http://www.cuhk.edu.hk/policy/academichonesty/ Signature Date Please attempt to solve all the problems as exercises. This coursework will not be graded. 2 1. Answer the following questions: (a) Assume that all radicands represent nonp-negative real numbers. Simplify each expression by removing as many as factors as possible from under the radical. √ p a2 b 3 3 8 9 (ii) (i) x 64x y z ; . ab−1 √ (b) Simplify m2 − 4m + 4, if possible. (Hint: Consider the cases m ≥ 2 and m < 2 separately.) (c) Assume all radicands are positive and denominators are non-zero. Simplify each expression by rationalizing the denominator: 3 √ ; 1+ 2 √ s−3 (iii) ; s−9 (i) 1 √ ; √ √r − √3 s+ s+3 √ (iv) √ . s− s+3 (ii) (d) Find the domain of each function: 1 ; x+4 7x (iii) f (x) = √ ; 4−x (i) f (x) = 1 f (x) = √ ; 3 x−1 x+4 . (iv) f (x) = 2 x + 3x + 2 (ii) 3 p (a)(i) x 3 64x3 y 8 z 9 = x(4xy 8/3 z 3 ) = 4x2 y 8/3 z 3 . √ a2 b (a)(ii) −1 = a2−1 b(1/2)−(−1) = ab3/2 . ab ( p m−2 if m ≥ 2; (b) m2 − 4m + 4 = (m − 2)2 = |m − 2| = −m + 2 if m < 2. √ √ √ 3(1 − 2) 3−3 2 3 √ = √ √ = 2 = −3 + 3 2. (c)(i) 1 −2 1+ 2 (1 + 2)(1 − 2) √ √ √ √ 1 r+ 3 r+ 3 √ = √ √ √ √ = (c)(ii) √ . r−3 r− 3 ( r − 3)( r + 3) √ s−3 (c)(iii) . The given denominator is already rationalized. s−9 √ Remark. If the question asked for rationalization of numerator, the answer would be √ √ 1 s−3 s−3 √ =√ . = √ s−9 ( s − 3)( s + 3) s+3 (c)(iv) √ √ √ √ √ √ s+ s+3 ( s + s + 3)2 ( s + s + 3)2 √ √ √ = √ = √ √ s − (s + 3) s− s+3 ( s − s + 3)( s + s + 3) √ √ 2 s + s + 3 + 2 s + 3s 2s + 3 + 2 s2 + 3s = =− . −3 3 4 (d)(i) (−∞, −4) ∪ (−4, ∞) or R \ {−4} 8 6 4 y 2 0 −2 −4 −6 −8 −8 −6 −4 −2 0 2 4 6 8 2 4 6 8 x (d)(ii) (−∞, 1) ∪ (1, ∞) or R \ {1} 8 6 4 y 2 0 −2 −4 −6 −8 −8 −6 −4 −2 0 x 5 (d)(iii) (−∞, 4) 200 150 y 100 50 0 −50 −20 −15 −10 −5 0 5 x (d)(iv) (−∞, −2) ∪ (−2, −1) ∪ (−1, ∞) or R \ {−1, −2}, since x2 + 3x + 2 = (x + 1)(x + 2). 50 40 30 20 y 10 0 −10 −20 −30 −40 −50 −4 −3 −2 −1 x 0 1 2 6 2. Answer the following questions: (a) Write each union as a single interval. (i) (−∞, −3) ∪ (−5, ∞); (ii) (−∞, −10] ∪ (−∞, −8]. (b) Write each intersection as a single interval. (i) (−∞, −3) ∩ [−5, ∞); (ii) [−20, −10] ∩ (−∞, −8]. (c) Write each set as an interval or as a union of two intervals. 1 ; (ii) {x| |2x − 3| ≥ 4} ; (i) x| |x − 4| < 10 2 (iii) {x| x + 4x > −3} . (a)(i) (−∞, ∞) = R, (a)(ii) (−∞, −8], since −3 > −5. since −8 > −10. (b)(i) [−5, −3), since −3 > −5. (b)(ii) [−20, −10], since −8 > −10. (c)(i) (3.9, 4.1), since |x − 4| < 1 ⇔ −0.1 < x − 4 < 0.1 ⇔ 3.9 < x < 4.1. 10 (c)(ii) (−∞, − 12 ] ∪ [ 72 , ∞), since |2x − 3| ≥ 4 ⇔ 2x − 3 ≥ 4 or 2x − 3 ≤ −4 ⇔ x ≥ 7 1 or x ≤ − . 2 2 (c)(iii) (−∞, −3) ∪ (−1, ∞), since x2 + 4x > −3 ⇔ (x + 3)(x + 1) > 0 ⇔ x < −3 or x > −1. 7 3. Consider the function ( 1, if x < 1; f (x) = 2x − 3, if x ≥ 1. (a) Sketch the graph of f (x) for −4 ≤ x ≤ 4. (b) Evaluate and simplify each of the following expressions. (i) f (1); (iii) f (1 + ) where > 0; (v) f (|x| + 2). (ii) f (−2); (iv) f (1 + ) where < 0; (a) 6 5 4 y 3 2 1 0 −1 −2 −4 −3 −2 −1 0 1 2 3 4 x (b)(i) f (1) = −1, since 2(1) − 3 = −1. (b)(ii) f (−2) = 1, since −2 < 1. (b)(iii) Since > 0, 1 + > 1, f (1 + ) = 2(1 + ) − 3 = 2 − 1. (b)(iv) Since < 0, 1 + < 1, f (1 + ) = 1. (b)(v) Since |x| + 2 > 1, f (|x| + 2) = 2(|x| + 2) − 3 = 2|x| + 1. 8 4. Recall that ( x, if x ≥ 0; |x| = −x, if x < 0. Graph the following functions: (a) f (x) = 3|2x + 6| − 4; (b) f (x) = |x2 − 3x + 2|; x−1 . (c) f (x) = |x + 1| (Hint : Sketch the graph of y = 2 x−1 =1− first.) x+1 x+1 (a) Let g(x) = |x|. Then, f (x) = 3g(2x + 6) − 4. So, we can obtain the graph of f (x) by performing the following transformations on the graph of g(x) : (a) move it to the left by 6 (b) horizontally scale it by 1 2 (c) vertically scale it by 3 (d) move it down by 4 10 8 6 4 y 2 0 −2 −4 −6 −8 −10 −10 −8 −6 −4 −2 0 2 4 6 8 10 x (Red indicates the part where the function is decreasing and blue indicates the part where the function is increasing.) 9 (b) One can first sketch the graph of y = x2 − 3x + 2 = (x − 1)(x − 2) Then reflect the part that y < 0 (i.e. 1 < x < 2) about the x-axis. 10 8 6 4 y 2 0 −2 −4 −6 −8 −10 −10 −8 −6 −4 −2 0 x 2 4 6 8 10 10 (c)Let us first sketch the graph of y = h(x) = x−1 2 =1− x+1 x+1 By letting g(x) = x1 , we have h(x) = −2g(x + 1) + 1 Therefore, its graph can be obtained by performing the following transformations on the graph of g(x) : (a) move it to the left by 1 (b) reflect it about the x-axis (c) vertically scale it by 2 (d) move it up by 1 The graph of f (x) can then be obtained by reflecting the part of h(x) that x < 1 about the x-axis. 10 8 6 4 y 2 0 −2 −4 −6 −8 −10 −10 −8 −6 −4 −2 0 x 2 4 6 8 10 11 5. Consider the function f (x) = x2 +1. In this exercise, we will estimate the area under the graph of f (x) for 0 ≤ x ≤ 3, that is, the area of the shaded region bounded by the curve y = f (x) and the lines y = 0, x = 0 and x = 3 in Figure 1. This can be done by approximating the region using rectangles. Figure 1: Graph of f (x). (a) In Figure 2(a), the horizontal interval [0,3] is divided into 3 equal segments. A rectangle is drawn above each of these segments as shown. What is the total area of the rectangles? (b) In Figure 2(b), the horizontal interval [0,3] is divided into 6 equal segments and 6 rectangles are drawn. What is the total area of these 6 rectangles? (a) 3 rectangles (b) 6 rectangles Figure 2: Area Estimate by rectangles. 12 (c) Which answer in part (a) and (b) is a better estimate of the area under the graph of f (x) for 0 ≤ x ≤ 3? Are the estimates greater or smaller than the actual area under the graph of f (x)? (d) (More difficult) Suppose the horizontal intervals [0,3] is divided into n equal segments and n rectangles are drawn similarly as in part (a) and (b). Show that the total area of these n rectangles is 9n(n − 1)(2n − 1) + 3. 2n3 The following identities may be useful: n(n + 1) 1 + 2 + ... + n = ; 2 n(n + 1)(2n + 1) 12 + 22 + . . . + n2 = . 6 (a.) The width of each rectangle is 1. The heights of the rectangles are f (0) = 1, f (1) = 2, f (2) = 5. Hence the total area is 1 · (1 + 2 + 5) = 8. (b.) The width of each rectangle is 0.5. The heights of the rectangles are f (0) = 1, f (0.5) = 1.25, f (1) = 2, f (1.5) = 3.25, f (2) = 5, f (2.5) = 7.25. Hence the total area is 0.5 · (1 + 1.25 + 2 + 3.25 + 5 + 7.25) = 9.875. (c.) Part (b) is a better estimation. The estimations are smaller than the actual area. 3 (d.) The width of each rectangle is . n The heights of the rectangles are 2 3i 9i2 3 f 0+ ·i = + 1 = 2 + 1, n n n where i = 0, 1, 2, ..., n − 1. Hence the total area is ! n−1 2 n−1 n−1 X X X 3 9i 3 9 · +1 = · i2 + 1 2 2 n i=0 n n n i=0 i=0 9(n − 1)(n)(2n − 1) 3 = · +n n 6n2 9n(n − 1)(2n − 1) = + 3. 2n3 13 6. Find the average rate of change of f over the interval [a, x]; that is f (x) − f (a) , x−a for each function. Be sure to simplify. (a) f (x) = x2 − 2x + 3; (b) f (x) = x3 ; 3 ; (c) f (x) = 1−x √ (d) f (x) = x + 4. (a.) f (x) − f (a) x2 − 2x + 3 − a2 + 2a − 3 (x − a)(x + a) − 2(x − a) = = x−a x−a x−a = x + a − 2. (b.) x 3 − a3 (x − a)(x2 + xa + a2 ) f (x) − f (a) = = = x2 + xa + a2 x−a x−a x−a (c.) 3(1−a)−3(1−x) (1−x)(1−a) 3 − 1−a = x−a 3 = (1 − a)(1 − x) f (x) − f (a) = x−a 3 1−x x−a = −3a + 3x (1 − a)(1 − x)(x − a) (d.) √ √ √ √ √ x+4− a−4 x− a x− a √ √ √ = = √ x−a x−a ( x − a)( x + a) 1 √ =√ x+ a f (x) − f (a) = x−a √ 14 7. Figure 3 shows the graphs of each of the six basic trigonometric functions. 1.5 1 1 y 0 0 −0.5 −1 −1 −1.5 −2pi −3pi/2 −pi −pi/2 0 pi/2 pi 3pi/2 −5pi/2 2pi −2pi −3pi/2 −pi y (a) y = sin (x) 5 4 4 3 3 2 2 1 1 y 0 −1 −2 −2 −3 −3 −4 −4 −pi/2 0 pi/2 pi −5 −2pi 3pi/2 −3pi/2 −pi −pi/2 (c) y = tan (x) 5 4 4 3 3 2 2 1 1 y 0 −1 −2 −2 −3 −3 −4 −4 0 2pi 5pi/2 0 pi/2 pi 3pi/2 2pi 0 −1 −pi/2 3pi/2 (d) y = cot (x) = 1/ tan (x) 5 −pi pi x (radians) x (radians) −5 −3pi/2 pi/2 0 −1 −pi 0 (b) y = cos (x) 5 −5 −3pi/2 −pi/2 x (radians) x (radians) y y 0.5 pi/2 pi x (radians) (e) y = sec (x) = 1/ cos (x) 3pi/2 −5 −2pi −3pi/2 −pi −pi/2 0 pi/2 pi 3pi/2 2pi x (radians) (f) y = cosec (x) = 1/ sin (x) Figure 3: Six trigonometric functions with restricted domains. The period of each trigonometric function is the length of each cycle, that is, the period is the horizontal distance between any point on the curve and the next corresponding point in the next cycle where the graph starts to repeat itself. The amplitude of the sine and cosine function is one-half the distance difference between the maximum and minimum values of the function. Here are graphing facts and relationships for the six trigonometric functions. 15 y y 2π : period = b y y = a sin (bx + c) amplitude = |a| = a cos (bx + c) amplitude = |a| = a sec (bx + c) = a cosec (bx + c) ( y = a tan (bx + c) π period = : b y = a cot (bx + c) Answer the following questions: (a) Graph y = −2 sin (12x) by hand; x 1 by hand; (b) Graph y = cos 3 4 π (c) Graph y = − sin 3x − by hand; 2 π by hand. (d) Graph y = tan 3x − 2 2π 2π π (a.) The amplitude is 2 and the period is p = = = . The effect of the negative b 12 6 sign is to reflect the curve about x-axis. 3 2 y 1 0 −1 −2 −3 −pi −3pi/4 −pi/2 −pi/4 0 x pi/4 pi/2 3pi/4 pi 16 (b) The amplitude is 2π 2π 1 and the period is = = 8π. 3 b 1/4 1.5 1 y 0.5 0 −0.5 −1 −1.5 −4pi −2pi 0 2pi 4pi x 2π c −π/2 π 2π = . The phase shift is = =− , (c) The amplitude is 1 and the period is b 3 b 3 6 π i.e. to the right. The effect of the negative sign is to reflect the curve about the x-axis. 6 1.5 1 y 0.5 0 −0.5 −1 −1.5 −2pi 0 x 2pi 17 (d) The period is π π c −π/2 π π = and the phase shift is = = − , i.e. to the right. b 3 b 3 6 6 The asymptotes are x= 1 π π π (n + 1)π 1 nπ + − c = nπ + + = b 2 3 2 2 3 we can simply write as x = nπ , n ∈ Z. 3 20 15 10 y 5 0 −5 −10 −15 −20 −pi/2 −pi/3 −pi/6 0 x pi/6 pi/3 pi/2 18 8. (Optional) Let us study the so-called cardinal sine function. The standard sinc x function without normalization is defined by sin x x 6= 0; sinc x = x 1 x = 0, as illustrated in Figure 4. 1.5 1 y 0.5 0 −0.5 −1 −1.5 −5pi −4pi −3pi −2pi −pi 0 pi 2pi 3pi 4pi 5pi x Figure 4: The standard sinc function without normization. Find the zero crossing points of the normalized sinc function: sin πx x 6= 0; sinc x = πx 1 x = 0. Inspection of Figure 4 shows that the sinc function which has not been normalized has the same zero crossing points as sin x, with exception of the crossing x = 0, that is at x = ±π, ±2π, ±3π, · · · , ±nπ, · · · . The normalized sinc function is 0 when sin(πx) = 0 except for x = 0. Hence, sin(x) = 0 when x = · · · , −3π, −2π, −1π, 0, 1π, 2π, 3π, · · · and so sin(πx) = 0 when x = · · · , −3, −2, −1, 0, 1, 2, 3, · · · . 19 Noting that sinc(0) is defined to have a value of 1, then we see that sinc(x) = 0 when x = · · · , −3, −2, −1, 1, 2, 3, · · · or sinc(x) = 0 when x = ±n for n = 1, 2, 3, · · · The plot of the normalized sinc function is shown below. 1.5 y 1 0.5 0 −0.5 −15 −10 −5 0 x 5 10 15