Calculus Preparation & Placement Evaluation, 2009 Department of Mathematics Wilfrid Laurier University Solutions to Sample Evaluation Simplifying Expressions Question 1 50 25 1 5 2 +9 ( 1) = 54 ************************************************************************ Recall order of operations: 50 1 5 25 = 50 = 50 = 50 = 54 B E D M A S brackets (work from inside out) exponents (which include roots) division/multiplication (in order, left to right) addition/subtraction (in order, left to right) 2 +9 25 15 +9 | {z } ( 1) | {z } 1 5+9 1 | {z } 5+9 ************************************************************************ ************************************************************************ Question 2 3 4 2 + 4 5 3 127 = 8 40 ************************************************************************ 3 4 2 + 4 5 3 8 = 11 4 + 4 5 = 110 32 + 40 40 = 127 40 3 8 (changing the mixed fraction to an improper fraction) 15 40 ( nding a common denominator for all terms) ************************************************************************ ************************************************************************ 1 Question 3 p 3 is an example of a number from which set? irrational numbers ************************************************************************ natural numbers ( N ) : f1; 2; 3; 4; 5; :::g integers ( Z ) : f:::; 4; 3; 2; 1; 0; 1; 2; 3; 4; :::g na o rational numbers ( Q ) : j a; b 2 Z, b 6= 0 b ! any value which can be represented as a terminating or repeating decimal irrational numbers: all non-terminating, non-repeating decimal values reals ( R ) : union of the rationals and irrationals p 3 = 1:732 050 808 ::: a non-terminating, non-repeating decimal value; p thus, 3 is an irrational number. ************************************************************************ ************************************************************************ Question 4 4 2 1 = 3+3 1 9 32 ************************************************************************ 4 2 1 3+3 1 = 1 42 3+ = = = = = 1 1 3 1 16 9 + 3 15 16 10 3 15 16 3 16 9 32 16 16 1 3 3 10 3 2 ************************************************************************ ************************************************************************ 2 Question 5 2 64x1=2 y 9 2=3 = 1 8x1=3 y 6 ************************************************************************ 2 64x1=2 y 9 2=3 2 = 2=3 64x1=2 y 9 2 = 2=3 x1=2 (64) 2=3 (y 9 ) 2 = = 2=3 p 3 2 64 x1=3 (y 6 ) 2 2 (4) x1=3 (y 6 ) 2 16x1=3 y 6 1 = 1=3 6 8x y = ************************************************************************ ************************************************************************ Question 6 If x 6= 1, then 2 (x + 3) x 4 (x 1 1) = 10 x 2x 1 ************************************************************************ 2 (x + 3) x 4 (x 1 1) 2x + 6 4x + 4 x 1 2x + 10 = x 1 10 2x = x 1 = ************************************************************************ ************************************************************************ Question 7 If x > 3, then 2 (x 1=2 3) x (x x 3 3) 1=2 x 6 =q 3 (x 3) ************************************************************************ 3 1=2 2 (x 3) x (x x 3 3) 1=2 = = = = (x 3) 1=2 x (x 3) 1=2 x 1=2 3) x 6 (x 3) 3=2 3) [2x 3 6 6 (x 1 x (x [2 (x 3 x] x] 3) x 6 =q 3 (x 3) ************************************************************************ ************************************************************************ Question 8 If x 6= 1, 2, then x+2 x4 16 x+1 2 = 2x 4 (x + 1) (x2 + 4) ************************************************************************ x+2 x4 16 x+1 = 2 2x 4 (x = = = = x+2 4) (x2 + 4) x+1 2 (x 2) (x x+2 2) (x + 2) (x2 + 4) x+1 2 (x 2) (x (x + 2) 2) (x + 2) (x2 + 4) 2 (x 2) (x + 1) (x2 1 + 4) 2 (x + 1) 2 (x + 1) (x2 + 4) ************************************************************************ ************************************************************************ Question 9 If x 6= 3, then 4+x x x2 = 4x + 3 1 x 12 x2 ************************************************************************ 12 x2 x2 + x 12 x2 4x + 3 (x + 4) (x 3) = (x 1) (x 3) x+4 = x 1 x+4 = 1 x x x2 = 4x + 3 ************************************************************************ ************************************************************************ 4 Question 10 The remainder, when 9 + 3x + 4x3 2x4 is divided by x 2, is: 15 ************************************************************************ Using the Factor/Remainder Theorem, we substitute x = 2 into the expression: 9 + 3 (2) + 4 (2) 3 2 (2) 4 = 9 + 32) + 4 (8) = 9 + 6 + 32 2 (16) 32 = 15 OR Using long division of polynomials: x Thus, 2 p 2x3 + 0x2 + 0x + 3 2x4 + 4x3 + 0x2 + 3x + 9 2x4 + 4x3 0x3 + 0x2 + 3x + 9 (3x 6) 15 remainder 2x4 + 4x3 + 3x + 9 = x 2 2x3 + 3 + 15 x 2 . ************************************************************************ ************************************************************************ Question 11 1 x+h+1 h 1 x+1 = 1 (x + 1) (x + h + 1) ************************************************************************ 1 x+h+1 h 1 x+1 = 1 1 h x+h+1 1 x+1 = 1 1 (x + 1) h (x + h + 1) (x + 1) = 1 (x + 1) (x + h + 1) h (x + 1) (x + h + 1) = 1 x+1 x h 1 h (x + 1) (x + h + 1) = 1 h h (x + 1) (x + h + 1) = 1 (x + 1) (x + h + 1) 1 (x + h + 1) (x + 1) (x + h + 1) ************************************************************************ ************************************************************************ 5 Question 12 9 , then 2 3 If x > 2x p = 2x + 9 3+ p 2x + 9 ************************************************************************ p p 2x 3 + 2x + 9 3 + 2x + 9 2x p p = 9 (2x + 9) 3 2x + 9 3 + 2x + 9 p 2x 3 + 2x + 9 = 9 2x 9 p 2x 3 + 2x + 9 = 2x p = 3 + 2x + 9 ************************************************************************ ************************************************************************ Question 13 Which of the following statements are true? b = b Answer: I only I. a 1+ a b II. m =0 0 III. p x2 9=x 3 2 IV. (q + 2) = q 2 + 4 ************************************************************************ I. a b = a b b a = b b 1= 1+ b m m II. is unde ned [ 6= 0 ] 0 0 p III. x2 9 cannot be simpli ed [ a X b p x2 9 6= x 3] 2 2 IV. (q + 2) = (q + 2) (q + 2) = q 2 + 2q + 2q + 4 = q 2 + 4q + 4 [ (q + 2) 6= q 2 + 4 ] ************************************************************************ ************************************************************************ Question 14 The expression x2 6x + 2, upon completing the square, is the same as: (x 2 3) 7 ************************************************************************ x2 6x + 2 = x2 | = (x = (x 6x + 9 9 + 2 {z } 2 9+2 2 7 3) 3) ************************************************************************ ************************************************************************ 6 Question 15 5 In the expansion of (2x + 1) , the coefficient of the x3 term is: 80 ************************************************************************ Binomial expansion: n (x + y) = xn + n n x 1 1 y+ n n x 2 2 2 y + + n n x k k k y + + n n 1 xy n 1 + yn 5 Thus, for (2x + 1) , the coef cient of the x3 term corresponds to n = 5, k = 2: 5 5 (2x) 2 2 2 (1) 5! 3 2 (2x) (1) 2! (5 2)! 120 = 8x3 (1) (2) (6) = [Recall: m! = m (m 1) (m = 10 8x3 ! thus, the required coef cient is 80. = 80x3 ************************************************************************ ************************************************************************ Solving Equations and Inequalities Question 16 The equation 9 (x + 8) = 2x has solution: x = (4 x) 38 3 ************************************************************************ 9x 9 (x + 8) = 2x (4 9x + 72 = 2x 4+x x = 4 72 6x = 76 x = 2x = x) 76 6 38 3 ************************************************************************ ************************************************************************ 7 2) (2) (1) ] Question 17 The possible values of x for which (2x + 1) (x 7 are: 4) = x= 1 ,3 2 ************************************************************************ (2x + 1) (x 2x2 2 2x 7 4 = 7 8x + x 4+7 =0 2x2 7x + 3 =0 2 ( 7) ( 7) = = 8x + x Using the quadraticq formula: x 4) Or using factoring techniques: 4 (2) (3) 2x2 2 (2) p (2x 49 24 4 p 25 7 = 4 7 5 7+5 , = 4 4 1 = , 3 2 = 7 2x 7x + 3 1) (x 1 x 3) =0 1 = 2 =0 =0 OR x 3 =0 x =3 ************************************************************************ ************************************************************************ Question 18 For which value(s) of m would the equation x2 6x + m = 0 have no real solutions? Answer: m > 9 ************************************************************************ A quadratic [ ax2 + bx + c ] will have two real roots when its discriminant [ b2 4ac ] is greater than zero; one real root when its discriminant is equal to zero; and no real roots when its less than zero. Thus, solve 2 ( 6) 4 (1) (m) <0 36 4m <0 4m < 36 m > 36 4 m >9 ************************************************************************ ************************************************************************ 8 Question 19 The possible value(s) of x for which p x are: x = x2 + 9 = 1 4 ************************************************************************ p x2 + 9 =1 Checking: x x2 + 9 = (1 x2 + 9 =1 2 x) p LS = RS =1 16 + 9 = p 25 = 5 ( 4) = 1 + 4 = 5 = LS X 2x + x2 2x = 8 x = 4 ************************************************************************ ************************************************************************ Question 20 Solve the system of equations: x + 5y 5x + 20y Answer: x = =4 = 13 3, y = 7 5 ************************************************************************ Using the method of elimination: x + 5y 5x + 20y 1 =4 = 13 2 (5) ! 5x + 25y = 20 3 5x + 20y = 13 2 0x + 5y =7 subtract y Substitute y = 7 into 1 : 5 x+5 7 5 =4 x+7 =4 x = Thus the system has the unique solution (x; y) = = 4 7 5 . 3 3; 7 . 5 ************************************************************************ ************************************************************************ 9 Question 21 A rectangle has a perimeter of 40 cm. If the length is 2 cm longer than twice the width, nd the area of the rectangle. Answer: 84 cm2 ************************************************************************ Let w be the width of the rectangle and l its length. ) l= P = 40 = 2w + 2l 40 2w 2 = 20 w l+w = 20 1 2w 3w w =2 = 18 =6 2 and l = 2w + 2 Using the method of elimination: l subtract Substitute w = 6 into 1 : = 20 . l+6 l = 14 Thus the rectangle has width of 6 cm, length of 14 cm and thus an area of 6 14 = 84 cm2 . ************************************************************************ ************************************************************************ Question 22 The set fx 2 R j x 0 or 6<x< 2g can be expressed in interval notation as: ( 6; 2) [ [0; 1) ************************************************************************ When using interval notation, a round bracket means that the endpoint is not included in the set while a square bracket indicates that the endpoint is included. fx 2 R j x x 6<x< 0 or 6<x< 2g can be represented on a real number line as: ———–> 0 2 1 ———— 6 2 0 1 Thus, the set can be represented by ( 6; 2) [ [0; 1). ************************************************************************ ************************************************************************ 10 Question 23 The solution of x2 2x 0 is the set of all x such that: x 8 2 or x 4 ************************************************************************ x2 (x (x 2x 8 0 4) (x + 2) 0 + 4) (x + 2) 1 (x 4) (x + 2) 2 0 2x 1 4 ———–> <———– 1 Therefore, x2 + 2 0 when x 8 1 4 2 or x 4. ! using interval notation, the solution set is ( 1; 2] [ [4; 1) ************************************************************************ ************************************************************************ Question 24 Find all possible solutions of the equation j5x 2j = 4x. 2 Answer: x = , 2 9 ************************************************************************ Note: If jaj = b then either a = b or a = Case 1: Case 2: 5x 5x 2 = 4x x =2 2 = 9x x Check: b. LS = j5 (2) 2j = j10 RS = 4 (2) = 8 = LS X 4x Check: =2 2 = 9 Therefore, the solution set is LS = 5 2 9 RS = 4 2 9 2 = = 2j = j8j = 8 10 9 18 = 9 8 = LS X 9 2 ;2 . 9 ************************************************************************ ************************************************************************ 11 8 8 = 9 9 Question 25 If jaj < jbj, then which of the following is true? Answer: b < a < b or b < a < b ************************************************************************ If b > 0, then jbj = b and jaj < b means If b < 0, then jbj = b and jaj < b < a < b. b means b < a < b. If b = 0, then jbj = 0 and jaj < 0 has no solution. Therefore, if jaj < jbj, then b < a < b or b < a < b. ************************************************************************ ************************************************************************ Functions and Graphing Question 26 If f (x) = x2 + 3x then f (h + 6) = h2 + 15h + 54 ************************************************************************ 2 f (h + 6) = (h + 6) + 3 (h + 6) = h2 + 12h + 36 + 3h + 18 = h2 + 15h + 54 ************************************************************************ ************************************************************************ Question 27 Which of the following represents a function? Answer: f( 9 5) , ( 3; 5) , (0; 5) , (3; 5)g ************************************************************************ A function maps each element of its domain to a unique range value. f( 9; 4) ; ( 9; 8) ; ( 3; 3) ; (0; 5)g ! f ( 9) = 4 and f ( 9) = ) this is not a function 8 f( 9; 5) ; ( 3; 5) ; (0; 5) ; (3; 5)g ! represents a function X f( 3; 9) ; ( 3; 3) ; ( 3; 0) ; ( 3; 6)g ! not a function, as the domain value 3 is mapped to 4 different range values f( 9; 4) ; (0; 5) ; (0; 5) ; (9; 4)g ! not a function as f (0) = 5 and f (0) = 5 ************************************************************************ ************************************************************************ 12 Question 28 1 is the 4 x2 the set of all x such that: 2<x<2 The domain of f (x) = p ************************************************************************ We cannot divide by zero nor take the square root of a negative value. Therefore, the domain of f is the set of x such that: x2 >0 x) (2 + x) >0 4 (2 (2 – x) (2 + x) (2 2 1 x2 > 0 when 1 2 ————– 2 2 x) (2 + x) > 0 Therefore, 4 – + 1 1 2 < x < 2. ************************************************************************ ************************************************************************ Question 29 The function f (x) = x2 1 can be defined by: ( x2 1 if x 1 or x 1 f (x) = 1 x2 if 1<x<1 ************************************************************************ f (x) = ( x2 (x if x2 if x2 x2 1 x2 1 1 = (x 1) (x + 1) 1) (x + 1) – + 1 Therefore, f (x) = 1 0 1<0 ( 1 + 1 1 x2 1 if x 1 or x 1 x2 if 1<x<1 1 Note, the equality can be added to either piece of(the function, but not both; x2 1 if x < 1 or x > 1 i.e., a valid answer would also be f (x) = 1 x2 if 1 x 1 ************************************************************************ ************************************************************************ 13 Question 30 The function h (x) = x2 + 2x f (x) = x2=3 2=3 can be thought of as the composition f g, where: g (x) = x2 + 2x ************************************************************************ (f g) (x) = f (g (x)) Thus, taking f (x) = x2=3 and g (x) = x2 + 2x, (f g) (x) = f x2 + 2x = x2 + 2x 2=3 = h (x) NOTE: If f (x) = x2 + 2x, g (x) = x2=3 then: (f g) (x) If f (x) = x2 + 2x then: (f = f x2=3 2=3 2 =x =f + 2 x2=3 = x 4=3 g) (x) g (x) = + 2x = If f (x) = x4=3 then: (f 2 3 = 2=3 2=3 2=3 = f (2x) 2=3 = (2x) 2 2=3 + 2 (2x) = 24=3 x4=3 + 25=3 x2=3 ************************************************************************ ************************************************************************ Question 31 3 The inverse function of y = (x + 4) + 8 is: p 3 y= x 8 4 ************************************************************************ 3 original function: y = (x + 4) + 8 inverse function: x = (y + 4) + 8 p 3 3 3 x 8 = (y + 4) x 8 =y+4 p 3 = x 8 y 2 +2 4 4 16 + = 9 3 9 g (x) = (2x) g) (x) 2 3 2 3 4 ************************************************************************ ************************************************************************ 14 2 3 Question 32 Which of the following is not a polynomial? Answer: f (x) = x 2 +4 ************************************************************************ A polynomial has the form: an xn + an n 1 1x + an n 2 2x where n 2 N and ai 2 R for 0 i + a2 x2 + a1 x1 + a0 + n. Thus: A. x2 B. C. x p D. x3 4 + 6x is a polynomial with n = 2, a2 = 1, a1 = 6, a0 = 2 4 + 4 is not a polynomial [ n = 22 = N] p 3 is a polynomial with n = 0, a0 = 3 4 is a polynomial with n = 3, a3 = 1, a2 = a1 = 0, a0 = 4 ************************************************************************ ************************************************************************ Question 33 If the graph of y = f (x) is obtained from the graph of y = x4 by shifting it one unit down and two units to the right, then the formula for the function y = f (x) is: 4 y = (x 2) 1 ************************************************************************ Transformations and Translations: Consider the graph of the function f (x). Then, if c > 0, f (x) c f (x c) is the graph of f shifted up [+] or down [ ] by c units is the graph of f shifted to the left [+] or to the right [ ] by c units cf (x) is the graph of f stretched [ if c > 1 ] or compressed [ if c < 1 ] vertically by a factor of c f (cx) is the graph of f compressed [ if c > 1 )] or stretched [ if c < 1 ] horizontally by a factor of c f (x) is the graph of f reflected in the x-axis f ( x) is the graph of f reflected in the y-axis ! f (x) shifted one unit down is represented by f (x) 1 ! f (x) 1 shifted two units to the right is represented by f (x Thus, y = (x 4 2) 2) 1 1 ************************************************************************ ************************************************************************ 15 Question 34 The relationship between the graphs of y = f (x) and y = f (x) is: a re ection in the x-axis ************************************************************************ See solution for Question 33. ************************************************************************ ************************************************************************ Question 35 Graph the region represented by the solution of 6x 7y 42. Answer: ************************************************************************ First, we consider the line 6x (i) 2 points on the line 7y = 42. We either need: ! x-intercept: 6x ! y-intercept: 6 (0) or (ii) 1 point and the slope of the line 10 42 , 7y = 42 , x = 7 42 , y = 6 6 6x + 42 = x+6 7 7 6 ! y-intercept: y = 6 and slope: m = 7 ! 6x LS = 6 (0) 10 -10 7y = 42 , 6x = 7y = 42 , y = Then determine which side of the line satis es the inequality by using a test point, say (0; 0) : Graph the line: -10 7 (0) = 7 (0) = 0 ? 42 = RS This is a true statement, thus the point (0; 0) is part of the solution and we shade the region below the line to represent all solutions. ************************************************************************ ************************************************************************ 16 Geometry Question 36 The distance between the points P (3; 7) and Q (6; 9) is: p 13 ************************************************************************ The distance between the points (x1 ; y1 ) and (x2 ; y2 ) is given by: q 2 2 d = (x2 x1 ) + (y2 y1 ) Thus, the distance between the points P (3; 7) and Q (6; 9): q p p 2 2 d = (6 3) + (9 7) = 9 + 4 = 13 ************************************************************************ ************************************************************************ Question 37 An equation of the line with a slope of 3 and passing through the point (2; 1) is: 3x y 7=0 ************************************************************************ A line with slope m passing through point (x0 ; y0 ) will have equation: y x y0 =m x0 Thus, a line with slope 3 passing through point (2; 1) has equation: y+1 x 2 =3 y+1 = 3 (x 0 2) = 3x 6 = 3x y 7 ************************************************************************ ************************************************************************ 17 Question 38 The vertex of the parabola given by f (x) = 4x2 8x 2 is: (1; 6) ************************************************************************ We “complete the square” to write the equation of the parabola in standard form: f (x) = 4x2 = 4 x2 h = 4 (x = 4 (x 8x 2 2x 2 1) 2 1) 2 i 1 2 6 Thus, the parabola has its vertex located at (1; 6). ************************************************************************ ************************************************************************ Question 39 Which of the following represent lines: I. y = 3 II. xy = 3 III. x2 + y 2 = 3 IV. x + y = 3 Answer: I and IV ************************************************************************ Horizontal lines (slope = 0) have general equation: y = c, c 2 R. Vertical lines (slope is unde ned) have general equation: x = k, k 2 R. Lines with slope = m have general equation: y = mx + b, m; b 2 R. I. y = 3 ! is a horizontal line X II. xy = 3 ! not a line III. x2 + y 2 = 3 ! not a line (actually this represents a circle) IV. x + y = 3 [or y = x + 3] ! a line with slope m = 1 X ************************************************************************ ************************************************************************ 18 Question 40 A circle centered at (0; 0) with radius 1 has equation: x2 + y 2 = 1 ************************************************************************ The general equation of a circle centered at (p; q) with radius r 2 has equation: (x 2 q) = r2 . p) + (y Thus, a circle centered at (0; 0) with radius 1 has equation: 2 (x 0) + (y 2 0) = (1) x2 + y 2 2 =1 Note: A circle centred at the origin, with radius 1, is referred to as the unit circle. ************************************************************************ ************************************************************************ Exponential and Logarithmic Functions Question 41 The expression log3 1 27 evaluates to: 3 ************************************************************************ loga b represents “the exponent which is applied to base a to obtain b” Letting x = log3 1 27 is equivalent to the equation 3x = By inspection, we nd that x = Thus, log3 1 27 = 3 as 3 3 = 1 3 3 = 1 . 27 1 . 27 3. ************************************************************************ ************************************************************************ 19 Question 42 Considering the domain x 2 ( 3; 0) [ (3; 1), the expression log log (x + 3) + log jx 3j x2 9 x3 is equivalent to: 3 log jxj ************************************************************************ Properties of logarithms include: loga (b c) loga = loga b + loga c b c = loga b loga (bc ) log loga c = c loga b x2 9 x3 = log (x 3) (x + 3) x3 = log (x 3) + log jx + 3j log x3 = log (x 3) + log jx + 3j 3 log jxj ************************************************************************ ************************************************************************ Question 43 The equation log5 (2x 3) = 0 has solution: x = 2 ************************************************************************ First note that logarithms are only de ned for positive values. Thus, for the given equation 3 . 2 We keep this in mind while solving the equation. 2x 3 > 0 () 2x > 3 () x > log5 (2x 2x 3) 3 2x x =0 = 50 = 1 =4 = 2 which is > 3 . 2 Thus, the equation has one solution, x = 2. ************************************************************************ ************************************************************************ 20 Question 44 The equation 4x = 8x has solution: x = 15 5 ************************************************************************ 4x = 8x OR 5 x = 23 x 5 22x = 23x 15 2x = 3x 22 x = x = 15 4x ln (4x ) x ln 4 15 15 = 8x 5 = ln 8x 5 x ln 4 = (x x ln 8 = x = 5 ln 8 ln 4 ln 8 = 5 ln 23 ln 22 ln 23 = 15 ln 2 2 ln 2 3 ln 2 = 5) ln 8 5 ln 8 15 ln 2 ln 2 = 15 ************************************************************************ ************************************************************************ Question 45 The function f (x) = 2x+3 f 1 1 has an inverse function given by: (x) = log2 (x + 1) 3 ************************************************************************ original function: y = 2x+3 1 inverse function: x = 2y+3 1 x+1 = 2y+3 log2 (x + 1) y =y+3 = log2 (x + 1) 3 ************************************************************************ ************************************************************************ 21 Trigonometry Question 46 3 radians 4 Convert 135o to radian measure: ************************************************************************ 1 radian = 180 o 1o = OR Therefore, 135o = 135 180 = radians 180 135 180 = 27 3 = radians 36 4 ************************************************************************ ************************************************************************ Question 47 5 . The angle is in radians. 6 p 3 Answer: 2 Evaluate cos ************************************************************************ The angle 5 6 radians falls within the second quadrant, so the CAST Rule tells us that cos Thus, cos 5 6 = cos 6 = 5 6 will be negative. p 3 . 2 ************************************************************************ ************************************************************************ 22 Question 48 Find the value of c in the given right triangle if B = radians and a = 6 cm. 3 Answer: c = 12 cm ************************************************************************ In relation to angle B, c is the length of the hypotenuse side and a the length of the adjacent side. Thus, we have cos B cos 3 c adjacent a = hypotenuse c 6 = c 6 = cos 3 = = 6 1 2 = 12 Therefore, the length of the hypotenuse is 12 cm. ************************************************************************ ************************************************************************ Question 49 Find the value of Angle C in the given triangle p if B = radians, b = 3 cm and c = 3 2 cm. 6 Answer: C = 4 radians ************************************************************************ Note that this is not a right triangle. Applying the sine law: sin C c sin C C sin B b c sin B = b p 3 2 sin 6 = 3 p 1 1 = 2 =p 2 2 = = 4 [as sin 1 =p ] 4 2 ************************************************************************ ************************************************************************ 23 Question 50 The function given by f ( ) = 3 sin (2 ) + 4 has period: ************************************************************************ Considering the general equation y = a sin (b ( + c)) + d: a = the amplitude of the sine wave 2 = the period of the sine wave b c = the phase shift of the sine wave d = a shift upwards [ d > 0 ] or downwards [ d < 0 ] of the wave. The period of f is then given by 2 = . 2 ************************************************************************ ************************************************************************ 24
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