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.

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]

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.