| Date | May 2009 | Marks available | 2 | Reference code | 09M.1.sl.TZ2.6 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Explain | Question number | 6 | Adapted from | N/A |
Question
A function f has its first derivative given by \(f'(x) = {(x - 3)^3}\) .
Find the second derivative.
Find \(f'(3)\) and \(f''(3)\) .
The point P on the graph of f has x-coordinate \(3\). Explain why P is not a point of inflexion.
Markscheme
METHOD 1
\(f''(x) = 3{(x - 3)^2}\) A2 N2
METHOD 2
attempt to expand \({(x - 3)^3}\) (M1)
e.g. \(f'(x) = {x^3} - 9{x^2} + 27x - 27\)
\(f''(x) = 3{x^2} - 18x + 27\) A1 N2
[2 marks]
\(f'(3) = 0\) , \(f''(3) = 0\) A1 N1
[1 mark]
METHOD 1
\({f''}\) does not change sign at P R1
evidence for this R1 N0
METHOD 2
\({f'}\) changes sign at P so P is a maximum/minimum (i.e. not inflexion) R1
evidence for this R1 N0
METHOD 3
finding \(f(x) = \frac{1}{4}{(x - 3)^4} + c\) and sketching this function R1
indicating minimum at \(x = 3\) R1 N0
[2 marks]
Examiners report
Many candidates completed parts (a) and (b) successfully.
Many candidates completed parts (a) and (b) successfully.
A rare few earned any marks in part (c) - most justifying the point of inflexion with the zero answers in part (b), not thinking that there is more to consider.
