IBDP Maths SL 5.8 Local maximum and minimum values AA HL Paper 1- Exam Style Questions- New Syllabus
Question
Let \( f(x) = \frac{4x^3}{3} – 16x \), defined for all real values of \( x \).
The curve given by \( y = f(x) \) has a local minimum at the point \((p, q)\), where \( p > 0 \).
Determine the values of \( p \) and \( q \).
Most-appropriate topic codes (IB Mathematics Analysis and Approaches 2021):
• SL 5.8 / AHL 5.12: Local maximum and minimum points; testing for maximum/minimum; optimization — whole question
• SL 5.3 / AHL 5.12: Derivative of \( f(x) = ax^n \); derivative from first principles — finding \( f'(x) \)
• SL 5.6 / AHL 5.15: Differentiation rules (sum, power, constant multiple) — differentiation step
• SL 5.3 / AHL 5.12: Derivative of \( f(x) = ax^n \); derivative from first principles — finding \( f'(x) \)
• SL 5.6 / AHL 5.15: Differentiation rules (sum, power, constant multiple) — differentiation step
▶️ Answer/Explanation
Step 1 — Find the derivative:
\( f'(x) = \frac{d}{dx} \left( \frac{4x^3}{3} – 16x \right) = 4x^2 – 16 \)
Step 2 — Set derivative equal to zero to find stationary points:
\( 4x^2 – 16 = 0 \)
\( x^2 = 4 \)
\( x = \pm 2 \)
Step 3 — Identify the positive critical point:
Given \( p > 0 \):
\( p = 2 \)
Step 4 — Find the corresponding \( y \)-coordinate \( q \):
\( q = f(2) = \frac{4(2)^3}{3} – 16(2) = \frac{32}{3} – 32 = \frac{32}{3} – \frac{96}{3} = -\frac{64}{3} \)
Final answers:
\( \boxed{p = 2}, \quad \boxed{q = -\frac{64}{3}} \)