(a) [3 marks]

\( f(x) = \frac{1}{3}x^3 + 2x^2 – 5x \).
\( f'(x) = x^2 + 4x – 5 \) (A1 for each term: \( x^2 \), \( 4x \), \( -5 \)).

(b) [4 marks]

Maximum at \( M \) when \( f'(x) = 0 \) (M1).
\( x^2 + 4x – 5 = 0 \implies (x + 5)(x – 1) = 0 \implies x = -5, 1 \) (A1).
Test second derivative: \( f”(x) = 2x + 4 \).
At \( x = -5 \): \( f”(-5) = 2(-5) + 4 = -6 < 0 \), maximum (A1).
\( x = -5 \) (A1).

(c) [3 marks]

Inflexion at \( N \) when \( f”(x) = 0 \) (M1).
\( f”(x) = 2x + 4 = 0 \implies x = -2 \) (A1).
Confirm by symmetry: midpoint of \( x = -5 \) and \( x = 1 \) is \( \frac{-5 + 1}{2} = -2 \) (A1).

(d) [4 marks]

Tangent at \( (3, 12) \): slope is \( f'(3) \) (M1).
\( f'(3) = 3^2 + 4 \cdot 3 – 5 = 16 \) (A1).
Equation: \( y – 12 = 16(x – 3) \implies y = 16x – 48 + 12 \) (M1).
\( y = 16x – 36 \) (A1).