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.