1. Exercise 7K Page 139 Questions 4, 5, 7, 12 & 13

2. Exercise 7K Page 139 Q4 (a) Factorise f(x) = x 3 + x 2 – 16x – 16 1 1 – 16 – 16 Try ± 1, ± 2, ± 4 4 4 5 1 20 4 16 f(x) = x 3 + x 2 – 16x – 16 = (x – 4)(1 x 2 + 5 x + 4) = (x – 4)(x + 4)(x + 1)

3. Exercise 7K Page 139 Q4 From 4(a) f(x) = x 3 + x 2 – 16x – 16 = (x – 4)(x + 4)(x + 1) 4(b) Write down the 4 points where f(x) cuts the axes. When x = 0 : f(0) = (0) 3 + (0) 2 – 16(0) – 16 = – 16 Thus cuts y axis at (0, – 16) Set f(x) = 0 to solve for roots of x when y = 0 (x – 4)(x + 4)(x + 1) = 0 Then x = 4 x = -4 x = -1 4 points are therefore (0, – 16), (4, 0), (-4, 0) & (-1, 0)

4. Exercise 7K Page 139 Q5 (x – 1) & (x + 4) are factors of f(x) = 2x 3 + 5x 2 + px + q (a) Find p & q. 2 5 p q 1 2 2 7 p+7 7p+q+7 = 0 2 5 p q -4 -8 2 12 p+12 -4p – 48 -3 -4p+q-48 = 0 If (x – 1) is a factor (x – 1) = 0 x = 1 is a solution If (x + 4) is a factor (x + 4) = 0 x = -4 is a solution

5. Exercise 7K Page 139 Q5 (x – 1) & (x + 4) are factors of f(x) = 2x 3 + 5x 2 + px + q (a) Find p & q. p + q + 7 = 0  -4p + q – 48 = 0  p + q = –7 (1) -4p + q = 48 (2) (1) – (2)  5p = –55 p = –11 Subst in to (1) to find q If p = -11 : p + q = –7 (1) -11 + q = –7 q = 4  p = -11 & q = 4

6. Exercise 7K Page 139 Q5 (b)Hence, solve f(x) = 0  Solve 2x 3 + 5x 2 -11x + 4 = 0 2 5 – 11 4 1 2 7 2 7 -4 f(x) = 2x 3 + 5x 2 – 11x + 4 = (x – 1)(2x 2 + 7x – 4 ) = 0 = (x – 1)(2x – 1 )(x + 4) = 0 Then x – 1 = 0 2x – 1 = 0 x + 4 = 0 2x = 1 x = 1 x = ½ x = – 4

7. Exercise 7K Page 139 Q7 f(x) = x 3 – 2x 2 – 5x + 6 & g(x) = x – 1 (a) Show f(g(x)) = x 3 – 5x 2 + 2x + 8 (x – 1) 2 = x 2 – 2x + 1 f(g(x)) = f(x – 1) = (x – 1) 3 – 2(x – 1) 2 – 5(x – 1) + 6 = (x 3 – 3x 2 + 3x – 1) – 2(x 2 – 2x + 1) – 5(x – 1) + 6 = x 3 – 3x 2 + 3x – 1 – 2x 2 + 4x – 2 – 5x + 5 + 6  Thus f(g(x)) = x 3 – 5x 2 + 2x + 8 as required. (x – 1) 3 = (x – 1)(x – 1) 2 = (x – 1)(x 2 – 2x + 1) = x 3 – 2x 2 + x – x 2 + 2x – 1 = x 3 – 3x 2 + 3x – 1

8. Exercise 7K Page 139 Q7 (b) Factorise fully f(g(x)) = x 3 – 5x 2 + 2x + 8 1 – 5 2 8 2 2 -4 1 -8 -3 -6 f(g(x)) = x 3 – 5x 2 + 2x + 8 = (x – 2)(1x 2 – 3x – 4 ) = (x – 2)(x – 4 )(x + 1)

9. Exercise 7K Page 139 Q7 (c)The function k is such that k(x) = 1. f(g(x)) For what values of x is the function not defined? From (b) f(g(x)) = x 3 – 5x 2 + 2x + 8 = (x – 2)(x – 4 )(x + 1) k(x) = 1 = 1  k(x) is undefined f(g(x)) (x – 2)(x – 4 )(x + 1) at x = 2, 4 & -1 As cannot have zero on the denominator

10. Exercise 7K Page 139 Q12 The diagram is a sketch of a graph of a cubic function y = f(x). If y = -16 is a tangent to the curve, find the formula for f(x). y = k(x – 0) 2 (x – 6) -16 = k(4 - 0) 2 (4 – 6) -16 = k(16)(– 2) -16 = – 32 k – 32 k = -16 k = ½ y = ½ (x – 0) 2 (x – 6) y = ½ x 2 (x – 6) y = ½ x 3 – 3x 2 y x o -16 4 6 y = f(x)

11. Exercise 7K Page 139 Q13 (a) Show that (x – 1) is a factor of f(x) = x 3 – 6x 2 + 9x – 4 1 -6 9 -4 1 1 1 -5 4 4 If (x – 1) is a factor (x – 1) = 0 x = 1 is a solution f (x) = x 3 – 6x 2 + 9x –4 = (x – 1)(1x 2 – 5x + 4 ) = (x – 1)(x – 1)(x – 4) = (x – 1) 2 (x – 4)

12. Exercise 7K Page 139 Q13 (b) Write down the co-ordinates where the graph y = f(x) meets the axes. f(x) = x 3 – 6x 2 + 9x – 4 = (x – 1) 2 (x – 4) x = 0 : f(0) = (0) 3 – 6(0) 2 + 9(0) – 4 = – 4  (0, -4) y = 0 f(x) = x 3 – 6x 2 + 9x – 4 = 0  (x – 1) 2 (x – 4) = 0 (x – 1) 2 = 0 (x – 4) = 0 x = 1 x = 4 3 points are therefore (0, – 4), (1, 0) & (4, 0)

13. Exercise 7K Page 139 Q13 (c) Find the stationary points of y = f(x) and determine their nature. f(x) = x 3 – 6x 2 + 9x – 4 For Stat Pts set f’(x) = 0 f’(x) = 3x 2 – 12x + 9 = 0 3(x 2 – 4x + 3) = 0 (x – 3)(x – 1) = 0  2 stat pts at x = 3 & x = 1 (x – 1) (x – 3) f’(x) Slope x  1  3  – – – 0 + – 0 + 0 + + 0 – 0 + x = 3 : f(3) = (3) 3 – 6(3) 2 + 9(3) – 4 = -4 x = 1 : f(1) = (1) 3 – 6(1) 2 + 9(1) – 4 = 0 Therefore Max (1, 0) & Min (3, -4)

14. Exercise 7K Page 139 Q13 (d) Sketch y = f(x) For f(x) = x 3 – 6x 2 + 9x – 4 Max (1, 0) & Min (3, -4) Cuts axis at (0, – 4), (1, 0) & (4, 0) y x o -4 4 1 y = f(x) Min (3, -4) Max (1, 0)

15. Exercise 7K Page 139 Q13 (e) Use the graph to find the number of solutions of the equation f(x) = -x f(x) = x 3 – 6x 2 + 9x – 4 f(x) = –x y x o -4 4 1 y = f(x) Min (3, -4) Max (1, 0) How many solutions are there when f(x) = –x is drawn on graph? Crosses at 3 points  3 solutions exist.