Chapter 2: Rational functions We are working with Cambridge Assessment International Education towards endorsement of this title. EXERCISE 2C ⇒ 4x 2 + 11x + 7 > 0
2x − 3 > 3, consider (2x − 3)(x + 2) > 3(x + 2) 2 x+2 So 2x 2 + x − 6 > 3x 2 + 12x + 12
1 For ⇒
x 2 + 11x + 18 < 0
⇒
(x + 9)(x + 2) < 0
⇒ (4x + 7)(x + 1) > 0 Hence x < − 7 or x > −1 4 You should make a quick sketch of your quadratic to see which values you need.
Hence −9 < x < −2 You don’t have to sketch curves in these cases, but it will aid you greatly, especially the consideration of any vertical asymptotes.
Then x 2 − xy − 3y = 0
Then xy − 3y = x 2 + x − 3, or x 2 + (1 − y)x + (3y − 3) = 0
FT
2 First let y(x + 3) = x 2, so that x 2 = xy + 3y
5 a First let y(x − 3) = x 2 + x − 3
The discriminant, b 2 − 4ac, is negative exactly when (1 − y) 2 − (4)(1)(3y − 3) < 0...................[1]
This has a negative discriminant exactly when (−y) 2 − 4(1)(−3y) < 0
Simplifying, we see that y 2 − 14y + 13 < 0 This factorises as (y − 1)(y − 13) < 0 which means 1 < y < 13 are the values that the curve cannot take.
Simplifying yields y 2 + 12y < 0, which factorises as y(y + 12) < 0
So −12 < y < 0 represents the values that cannot occur.
Again, a simple sketch will help you to check your solution.
A
Thus, y < −12 and y > 0 are the possibly values for y. You can also just consider b 2 − 4ac > 0 for the values you can have.
b By substituting y = 1 into [1], we obtain x 2 = 0, therefore (0, 1) is a turning point. Also from [1]: y = 13 ⇒ x 2 − 12x + 36 = 0 ⇒ (x − 6) 2 = 0
R
3 First determine the number of turning points. x 2 − 2x − 4 To do this, we differentiate y = using the 3x − 2 quotient rule.
So there is another turning point at (6, 13) 2 3x − 2 − 4x 2 , then as |x | → ∞, y → −4x 2 x − 2 − 3x 2 −3x So the horizontal asymptote is y = 4 3 Hence k = 4 3 b From the denominator, 3x 2 − x + 2 = 0
6 a If y =
dy (3x − 2)(2x − 2) − (x 2 − 2x − 4) .3 = dx (3x − 2) 2 2 2 6x − 10x + 4 − (3x − 6x − 12) 3x 2 − 4x + 16 = = (3x − 2) 2 (3x − 2) 2
D
So
( )
2 2 44 + Since the numerator is 3x 2 − 4x + 16 = 3 x − , 3 3 dy > 0. we have dx This tells us that the numerator is always positive, hence it is never zero.
Then b 2 − 4ac = 1 − 24 = −23. Since the value of the discriminant is < 0, there are no solutions. Hence there are no vertical asymptotes. c Let (x − 2 − 3x 2)y = 3x − 2 − 4x 2 Then (3y − 4)x 2 + (3 − y)x + (2y − 2) = 0
So there are no turning points.
This function has positive discriminant exactly when
Since there are also no horizontal asymptotes we conclude that all y values are possible. So y ∈R
(3 − y) 2 − 4(3y − 4)(2y − 2) > 0 ⇒ 9 − 6y + y 2 − 24y 2 + 56y − 32 > 0 ⇒ 23y 2 − 50y + 23 < 0
x−2 < 5 consider (x − 2)(x + 1) < 5(x + 1) 2 x+1 So x 2 − x − 2 < 5x 2 + 10x + 5
4 For
For the critical values, y= 13
50 ± √502 − 4 × 23 × 23 25 ± 4√6 = 2 × 23 23
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