Find $f''(x)$.

The graph of $f$ has a point of inflexion when $x = 1$.

Show that $k = 3$.

Find $f'( - 2)$.

Find the equation of the tangent to the curve of $f$ at $( - 2,{\text{ }}1)$, giving your answer in the form $y = ax + b$.

Given that $f'( - 1) = 0$, explain why the graph of $f$ has a local maximum when $x =  - 1$.

$f''(x) = 6x - 2k$     A1A1     N2

[2 marks]

substituting $x = 1$ into $f''$     (M1)

eg$\;\;\;f''(1),{\text{ }}6(1) - 2k$

recognizing $f''(x) = 0\;\;\;$(seen anywhere)     M1

correct equation     A1

eg$\;\;\;6 - 2k = 0$

$k = 3$     AG     N0

[3 marks]

correct substitution into $f'(x)$     (A1)

eg$\;\;\;3{( - 2)^2} - 6( - 2) - 9$

$f'( - 2) = 15$     A1     N2

[2 marks]

recognizing gradient value (may be seen in equation)     M1

eg$\;\;\;a = 15,{\text{ }}y = 15x + b$

attempt to substitute $( - 2,{\text{ }}1)$ into equation of a straight line     M1

eg$\;\;\;1 = 15( - 2) + b,{\text{ }}(y - 1) = m(x + 2),{\text{ }}(y + 2) = 15(x - 1)$

correct working     (A1)

eg$\;\;\;31 = b,{\text{ }}y = 15x + 30 + 1$

$y = 15x + 31$     A1     N2

[4 marks]

METHOD 1 (${{\text{2}}^{{\text{nd}}}}$ derivative)

recognizing $f'' < 0\;\;\;$(seen anywhere)     R1

substituting $x =  - 1$ into $f''$     (M1)

eg$\;\;\;f''( - 1),{\text{ }}6( - 1) - 6$

$f''( - 1) =  - 12$     A1

therefore the graph of $f$ has a local maximum when $x =  - 1$     AG     N0

METHOD 2 (${{\text{1}}^{{\text{st}}}}$ derivative)

recognizing change of sign of $f'(x)\;\;\;$(seen anywhere)     R1

eg$\;\;\;$sign chart$\;\;\;$

correct value of $f'$ for $- 1 < x < 3$     A1

eg$\;\;\;f'(0) =  - 9$

correct value of $f'$ for $x$ value to the left of $- 1$     A1

eg$\;\;\;f'( - 2) = 15$

therefore the graph of $f$ has a local maximum when $x =  - 1$     AG     N0

[3 marks]

Total [14 marks]

Well answered and candidates coped well with $k$ in the expression.

Mostly answered well with the common error being to substitute into $f'$ instead of $f''$.

A straightforward question that was typically answered correctly.

Some candidates recalculated the gradient, not realising this had already been found in part c). Many understood they were finding a linear equation but were hampered by arithmetic errors.

Using change of sign of the first derivative was the most common approach used with a sign chart or written explanation. However, few candidates then supported their approach by calculating suitable values for $f'(x)$. This was necessary because the question already identified a local maximum, hence candidates needed to explain why this was so. Some candidates did not mention the ‘first derivative’ just that ‘it’ was increasing/decreasing. Few candidates used the more efficient second derivative test.