CHAPTER 36 FUNCTIONS AND THEIR CURVES
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
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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
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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
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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:
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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:
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(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
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(b) A sketch of f(x) against x is shown below. Since the function is symmetrical about origin, it is an odd function
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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
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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
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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
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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
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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
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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
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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
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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:
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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:
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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:
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(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
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(b) A sketch of f(x) against x is shown below. Since the function is symmetrical about origin, it is an odd function
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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
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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
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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
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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
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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
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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
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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:
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© 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
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© 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
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© 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
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© 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
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