Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
CHAPTER 36 FUNCTIONS AND THEIR CURVES EXERCISE 147 Page 385 1. Sketch the following graph, showing relevant points: y = 3x – 5 2. Sketch the following graph, showing relevant points: y = – 3x + 4 3. Sketch the following graph, showing relevant points: y = x2 + 3 591 © 2014, John Bird 4. Sketch the following graph, showing relevant points: y = (x – 3)2 5. Sketch the following graph, showing relevant points: y = (x – 4)2 + 2 6. Sketch the following graph, showing relevant points: y = x – x2 7. Sketch the following graph, showing relevant points: y = x3 + 2 592 © 2014, John Bird 8. Sketch the following graph, showing relevant points: y = 1 + 2 cos 3x 9. Sketch the following graph, showing relevant points: y = 3 – 2 sin (x + π 4 ) 10. Sketch the following graph, showing relevant points: y = 2 ln x 593 © 2014, John Bird 594 © 2014, John Bird EXERCISE 148 Page 387 1. Determine whether the following functions are even, odd or neither even nor odd: (a) x4 (b) tan 3x (c) 2e3t (d) sin2 x (a) Let f(x) = x 4 . Since f(–x) = f(x) then x 4 is an even function and is symmetrical about the f(x) axis as shown below: (b) Let f(x) = tan 3x. Since f(–x) = – f(x) then tan 3x is an odd function and is symmetrical about the origin as shown below: (c) Let f(t) = 2 e3t . The function is neither even not odd, and is as shown below: (d) Let f(x) = sin 2 x . Since f(–x) = f(x) then sin 2 x is an even function and is symmetrical about the f(x) axis as shown below: 595 © 2014, John Bird 2. Determine whether the following functions are even, odd or neither even nor odd: (a) 5t3 (b) ex + e–x (c) cos θ θ (d) ex (a) Let f(x) =5t3. Since f(–x) = – f(x) then 5t3 is an odd function and is symmetrical about the origin as shown below: (b) Let f(x) = ex + e–x. Since f(–x) = f(x) then sin 2 x is an even function and is symmetrical about the f(x) axis as shown below: (c) Let f(t) = cos θ θ . Since f(–x) = – f(x) then cos θ θ is an odd function and is symmetrical about the origin as shown below: 596 © 2014, John Bird (d) Let f(x) = ex. The function is neither even not odd, and is as shown below: 3. State whether the following functions which are periodic of period 2π are even or odd: θ , when − π ≤ θ ≤ 0 (a) f(θ) = −θ , when 0 ≤ θ ≤ π π π x, when − 2 ≤ x ≤ 2 (b) f(x) = 0, when π ≤ x ≤ 3π 2 2 (a) A sketch of f(θ) against θ is shown below. Since the function is symmetrical about the f(θ) axis, it is an even function 597 © 2014, John Bird (b) A sketch of f(x) against x is shown below. Since the function is symmetrical about origin, it is an odd function 598 © 2014, John Bird EXERCISE 149 Page 389 1. Determine the inverse of: f(x) = x + 1 If y = f(x), then y=x+1 Transposing for x gives: x=y–1 Interchanging x and y gives: y=x–1 Hence, if f(x) = x + 1, then f −1 ( x)= x − 1 2. Determine the inverse of: f(x) = 5x – 1 If y = f(x), then y = 5x – 1 Transposing for x gives: x= y +1 5 Interchanging x and y gives: y= x +1 5 Hence, if f(x) = 5x – 1, then f −1 = ( x) 1 ( x + 1) 5 3. Determine the inverse of: f(x) = x3 + 1 If y = f(x), then y = x3 + 1 Transposing for x gives: x= 3 y −1 Interchanging x and y gives: y= 3 x −1 3 x −1 Hence, if f(x) = x3 + 1, then f −1 ( = x) 4. Determine the inverse of: f(x) = 1 +2 x If y = f(x), then y= 1 +2 x Transposing for x gives: x= 1 y−2 599 © 2014, John Bird Interchanging x and y gives: Hence, if f ( x)= y= 1 x−2 1 1 + 2 , then f −1 ( x) = x−2 x 5. Determine the principal value of the inverse function: sin −1 (– 1) Using a calculator (set on radians), sin −1 (– 1) = –1.5708 rad or – π 2 rad 6. Determine the principal value of the inverse function: cos −1 0.5 Using a calculator (set on radians), cos −1 0.5 = 1.0472 rad or π 3 rad 7. Determine the principal value of the inverse function: tan −1 1 Using a calculator (set on radians), tan −1 1 = 0.7854 rad or π 4 rad 8. Determine the principal value of the inverse function: cot −1 2 Using a calculator (set on radians), cot −1 2 = tan −1 1 = 0.4636 rad 2 9. Determine the principal value of the inverse function: cosec −1 2.5 Using a calculator (set on radians), cosec −1 2.5 = sin −1 1 = 0.4115 rad 2.5 10. Determine the principal value of the inverse function: sec −1 1.5 1 Using a calculator (set on radians), sec −1 1.5 = cos −1 = 0.8411 rad 1.5 600 © 2014, John Bird 1 11. Determine the principal value of the inverse function: sin −1 2 π 1 Using a calculator (set on radians), sin −1 = 0.7854 rad or 4 rad 2 12. Evaluate x, correct to 3 decimal places: x = sin −1 1 4 8 + cos −1 – tan −1 3 5 9 1 4 8 Using a calculator (set on radians), x = sin −1 + cos −1 − tan −1 3 5 9 = 0.3398 + 0.6435 – 0.7266 = 0.257 13. Evaluate y, correct to 4 significant figures: y = 3 sec −1 2 – 4 cosec −1 2 + 5 cot −1 2 Using a calculator (set on radians), y = 3sec −1 2 − 4 cosec −1 2 + 5cot −1 2 1 1 1 −1 = 3cos −1 + 5 tan −1 − 4sin 2 2 2 = 2.3562 – 3.1416 + 2.3182 = 1.533 601 © 2014, John Bird EXERCISE 150 Page 395 1. Determine the asymptotes parallel to the x- and y-axes for y = y= x−2 hence, x +1 x−2 x +1 y ( x + 1) = x − 2 and y ( x + 1) − x + 2 = 0 i.e. yx + y − x + 2 = 0 i.e. x ( y − 1) + y + 2 = 0 (1) (2) From equation (1), equating highest power of y to zero gives: x + 1 = 0, i.e. x = –1 From equation (2), equating highest power of x to zero gives: y – 1 = 0, i.e. y = 1 Hence, asymptotes parallel to the x- and y-axes occur at y = 1 and x = –1 2. Determine the asymptotes parallel to the x- and y-axes for y 2 = y2 = x hence, x −3 x x −3 y 2 ( x − 3) = x and y 2 ( x − 3) − x = 0 i.e. y2 x − 3y2 − x = 0 i.e. x ( y 2 − 1) − 3 y 2 = 0 (1) (2) From equation (1), equating highest power of y to zero gives: x – 3 = 0, i.e. x = 3 From equation (2), equating highest power of x to zero gives: y 2 – 1 = 0, i.e. y = ± 1 Hence, asymptotes parallel to the x- and y-axes occur at x = 3, y = 1 and y = –1 3. Determine the asymptotes parallel to the x- and y-axes for y = y= i.e. x( x + 3) ( x + 2)( x + 1) hence, y(x + 2)(x + 1) = x(x + 3) x( x + 3) ( x + 2)( x + 1) (1) y ( x 2 + 3x + 2 ) − x 2 − 3x = 0 602 © 2014, John Bird and yx 2 + 3 yx + 2 y − x 2 − 3 x = 0 i.e. x 2 ( y − 1) + 3 xy − 2 y − 3 x = 0 (2) From equation (1), equating highest power of y to zero gives: (x + 2)(x + 1) = 0, i.e. x = –2 and x = –1 From equation (2), equating highest power of x to zero gives: y – 1 = 0, i.e. y = 1 Hence, asymptotes parallel to the x- and y-axes occur at x = –2, x = –1 and at y = 1 4. Determine all the asymptotes for 8 x − 10 + x3 − xy 2 = 0 Substituting y = mx + c gives 8 x − 10 + x3 − x ( mx + c ) = 0 2 Simplifying gives 8 x − 10 + x3 − x ( m 2 x 2 + 2mxc + c 2 ) = 0 i.e. 8 x − 10 + x 3 − m 2 x 3 − 2mx 2 c − xc 2 = 0 and x 3 (1 – m 2 ) – 2mcx 2 – c 2 x + 8x – 10 = 0 Equating the coefficient of the highest power of x (i.e. x 3 in this case) gives 1 – m 2 = 0, from which, m=±1 Equating the coefficient of the next highest power of x (i.e. x 2 in this case) gives –2mc = 0, from which, c = 0 Hence y = mx + c = ± 1x + 0, i.e. y = x and y = –x are asymptotes To determine any asymptotes parallel to the x- and y-axes for the function 8 x − 10 + x3 − xy 2 = 0: Equating the coefficient of the highest power of x term to zero gives 1 = 0 which is not an equation of a line. Hence there is no asymptote parallel with the x-axis Equating the coefficient of the highest power of y term to zero gives –x = 0 from which, x = 0 Hence, x = 0, y = x and y = –x are asymptotes for the function 8 x − 10 + x3 − xy 2 = 0 5. Determine all the asymptotes for x 2 ( y 2 − 16 ) = y Equating highest power of x to zero gives: y 2 − 16 = 0, i.e. y = ±4 603 © 2014, John Bird Since x 2 ( y 2 − 16 ) = y then x 2 y 2 − 16 x 2 − y = 0 Equating highest power of y to zero gives: x 2 = 0, i.e. x = 0 Let y = mx + c, then 2 x 2 [ m 2 x 2 + 2mxc + c 2 − 16] = mx + c i.e. i.e. x 2 ( mx + c ) − 16 = mx + c m 2 x 4 + 2mcx3 + c 2 x 2 − 16 x 2 − mx − 1 =0 Equating coefficient of highest power of x to zero gives: m 2 = 0 , i.e. m = 0 Equating next coefficient of highest power of x to zero gives: 2mc = 0, i.e. c = 0 Hence, the only asymptotes occur at y = 4, y = –4 and at x = 0 6. Determine the asymptotes and sketch the curve for y = Since y = x2 − x − 4 then y(x + 1) = x 2 − x − 4 x +1 x2 − x − 4 x +1 (1) Equating the coefficient of the highest power of x term to zero gives 1 = 0 which is not an equation of a line. Hence there is no asymptote parallel with the x-axis. Equating the coefficient of the highest power of y term to zero gives x + 1 = 0 from which, x = –1 Substituting y = mx + c in equation (1) gives (mx + c)(x + 1) = x 2 − x − 4 Simplifying gives mx 2 + mx + cx + c = x 2 − x − 4 i.e. mx 2 + mx + cx + c – x 2 + x + 4 = 0 and x 2 (m – 1) + x(m + c + 1) + c + 4 = 0 Equating the coefficient of the highest power of x (i.e. x 2 in this case) gives m – 1 = 0, from which, m=1 Equating the coefficient of the next highest power of x (i.e. x in this case) gives m + c + 1 = 0, from which, c = –2 (since m = 1) Hence y = mx + c = x – 2, i.e. y = x – 2 is an asymptote 604 © 2014, John Bird Hence, x = –1, y = x – 2 are asymptotes for the function y = x2 − x − 4 x +1 The curve is shown sketched below 7. Determine the asymptotes and sketch the curve for xy 2 − x 2 y + 2 x − y = 5 5 xy 2 − x 2 y + 2 x − y = (1) Equating the highest power of y to zero gives: x = 0, which is an asymptote Equating the highest power of x to zero gives: –y = 0, i.e. y = 0, which is an asymptote Letting y = mx + c in equation (1) gives: x ( mx + c ) − x 2 ( mx + c ) + 2 x − (mx + c) = 5 2 i.e. x ( m 2 x 2 + 2mcx + c 2 ) − mx3 − cx 2 + 2 x − mx − c − 5 = 0 and m 2 x3 + 2mcx 2 + c 2 x − mx3 − cx 2 + 2 x − mx − c − 5 = 0 i.e. ( m2 − m ) x3 + ( 2mc − c ) x 2 + x ( c 2 + 2 − m ) − c − 5 =0 Equating the coefficient of the highest power of x to zero gives: m 2 − m = 0 , i.e. m(m – 1) = 0 i.e. m = 0 or m = 1 Equating the coefficient of the next highest power of x to zero gives: 2mc – c = 0 605 © 2014, John Bird When m = 0, c = 0 and when m = 1, 2c – c = 0, i.e. c = 0 Hence, y = mx + c becomes y = x, which is an asymptote. Thus, asymptotes occur at x = 0, y = 0 and at y = x A sketch of the curve xy 2 − x 2 y + 2 x − y =, 5 together with its asymptotes, is shown below: 606 © 2014, John Bird EXERCISE 151 Page 399 1. Sketch the graphs of (a) y = 3 x 2 + 9 x + (a) y = 3 x 2 + 9 x + 7 4 (b) y = −5 x 2 + 20 x + 50 dy = 6 x + 9 = 0 for a turning point dx x= − from which, 7 4 9 = –1.5 6 When x = –1.5, y = 3(−1.5) 2 + 9(−1.5) + 1.75 = –5 Hence, a turning point occurs at (–1.5, –5) d2 y = 6 , which is positive, hence, (–1.5, –5) is a minimum point d x2 At x = 0, y = 7 or 1.75 4 A sketch of the graph y = 3 x 2 + 9 x + (b) y = −5 x 2 + 20 x + 50 from which, 7 is shown below and is seen to be a parabola 4 dy 0 for a turning point = −10 x + 20 = dx 20 = 10x and x = 2 When x = 2, y = −5(2) 2 + 20(2) + 50 = 70 Hence, a turning point occurs at (2, 70) d2 y = −10 , which is negative, hence, (2, 70) is a maximum point d x2 At x = 0, y = 50 607 © 2014, John Bird A sketch of the graph y = −5 x 2 + 20 x + 50 is shown below y 2 2. Sketch the curve depicting: = x 4 1 − 4 y 2 Since= x 4 1 − 4 x then = 4 x2 y 2 or + = 1 16 16 from which, y 2 1 − 4 2 x y and = 1 − 4 4 x2 + y 2 = 16 or 2 x2 + y 2 = 42 This is the equation of a circle, centre (0, 0) and radius 4 units, as shown below 3. Sketch the curve depicting: Since x= y 9 then y=9 x x= and y 9 y 2 = 81x This is the equation of a parabola, symmetrical about the x-axis, vertex at (0, 0), as shown below 608 © 2014, John Bird 4. Sketch the curve depicting: y 2 = Since y 2 = x 2 − 16 then 4 x 2 − 16 4 2 4 y= x 2 − 16 i.e. 16 = x2 − 4 y 2 i.e. x2 4 y 2 − = 1 16 16 and x2 y 2 − = 1 which is a hyperbola, symmetrical about the x- and y42 22 axes, distance between vertices being 2(4), i.e. 8 units along the x-axis A sketch of y 2 = x 2 − 16 x2 y 2 , i.e. − = 1 is shown below 42 22 4 5. Sketch the curve depicting: Since and y2 x2 then = 5− 5 2 y2 x2 = 5− 5 2 x2 y 2 5 + = 2 5 x2 y 2 1 + = 10 25 609 © 2014, John Bird x2 i.e. ( 10 ) 2 + y2 = 1 52 which is an ellipse, centre (0, 0), major axis, AB = 2(5) = 10 units along the y-axis, and minor axis, CD = 2 10 along the x-axis. A sketch of the curve y2 x2 = 5 − , i.e. 5 2 x2 ( 10 ) 2 + y2 = 1 is shown below 52 6. Sketch the curve depicting: = x 3 1+ y2 Since = x 3 1+ y2 i.e. x then = 3 1 + y 2 and x2 x2 = 1 + y 2 or − y2 = 1 32 32 x2 y 2 − = 1 which is the equation of a hyperbola, symmetrical about the x- and y-axes, 32 12 with distance between vertices 2(3), i.e. 6 units along the x-axis, as shown below 7. Sketch the curve depicting: x 2 y 2 = 9 Since x 2 y 2 = 9 then y 2 = 9 x2 and y= 3 x 610 © 2014, John Bird which is a rectangular hyperbola, lying in the 1st and 3rd quadrants only, as shown in the sketch below. 8. Sketch the curve depicting: = x Since = x 1 3 ( 36 − 18 y 2 ) then = 3x x2 2 y 2 i.e. + = 1 4 4 ( 36 − 18 y 2 ) ( 36 − 18 y 2 ) x 2= 4 − 2 y 2 Dividing throughout by 9 gives: and 1 3 x2 y 2 + = 1 or 22 2 2 and 9 x= 36 − 18 y 2 from which, x 2 + 2 y 2 = 4 x2 + 22 y2 ( ) 2 = 1 2 This is the equation of an ellipse, centre (0, 0), major axis 2(2), i.e. 4 units along the x-axis, minor axis 2 2 units along the y-axis, as shown below 9. Sketch the circle given by the equation x 2 + y 2 − 4 x + 10 y + 25 = 0 Since x 2 + y 2 − 4 x + 10 y + 25 = 0 611 © 2014, John Bird then ( x − 2 ) + ( y + 5) 2 2 −4= 0 ( x − 2 ) + ( y + 5) 2 i.e. 2 = 22 which is a circle of centre (2, –5) and radius 2, as shown in the sketch below. 10. Describe the shape of the curve represented by the equation = y Since= y and 2 then y= 3 (1 − x 3 ) 2 3 (1 − x ) i.e. y 2 = 3 − 3x 2 y2 x2 + x2 = 1 i.e. + 3 12 y 2 + 3x 2 = 3 from which, 3 (1 − x 2 ) y2 ( 3) 2 = 1 This is the equation of an ellipse, centre (0, 0), minor axis 2 units along the x-axis, major axis 2 3 units along the y-axis 11. Describe the shape of the curve represented by the equation = y Since = y then = y 2 3 ( x 2 − 1) 3 ( x 2 − 1) and = y 2 3x 2 − 3 i.e. = 3 3x 2 − y 2 = 1 x2 − i.e. i.e. 3 ( x 2 − 1) x2 (1) 2 − y2 ( 3) 2 y2 3 = 1 612 © 2014, John Bird which is a hyperbola, symmetrical about the x- and y-axes, with vertices 2(1) = 2 units apart along the x-axis. 12. Describe the shape of the curve represented by the equation = y Since = y 9 − x2 9 − x2 32 9 or x 2 + y 2 = y 2= 9 − x 2 i.e. x 2 + y 2 = then This is the equation of a circle, centre (0, 0) and radius 3 units 13. Describe the shape of the curve represented by the equation y = 7 x −1 Since y = 7 x −1 then y= 7 x This is the equation of a rectangular hyperbola, lying in the 1st and 3rd quadrants, symmetrical about x- and y-axes 1 14. Describe the shape of the curve represented by the equation y = ( 3 x ) 2 1 Since y = ( 3 x ) 2 ( 3x ) then y = or y 2 = 3x which is a parabola, vertex at (0, 0) and symmetrical about the x-axis 15. Describe the shape of the curve represented by the equation y 2 − 8 =−2 x 2 Since y 2 − 8 =−2 x 2 then y 2 + 2 x2 = 8 and y 2 2x2 + = 1 8 8 i.e. y2 ( 8) 2 + x2 = 1 (2) 2 which is an ellipse, centre (0, 0), with major axis 2 8 units along the y-axis, and minor axis 2(2) = 4 units along the x-axis 613 © 2014, John Bird