IB Mathematics SL 5.3 Derivative of f(x) AA SL Paper 1- Exam Style Questions- New Syllabus

Step 1: Find the derivative
Differentiating \( f(x) = \frac{4}{3}x^3 – 16x \) using the power rule:
\( f'(x) = \frac{4}{3}(3x^2) – 16 \)
\( f'(x) = 4x^2 – 16 \)

Step 2: Find stationary points
Set \( f'(x) = 0 \) to locate turning points:
\( 4x^2 – 16 = 0 \)
\( 4(x^2 – 4) = 0 \)
\( x = \pm 2 \)

Step 3: Identify the local minimum
The question states the minimum is at \( p \) where \( p > 0 \).
Alternatively, checking the second derivative: \( f”(x) = 8x \).
At \( x = 2 \), \( f”(2) = 16 > 0 \) (Concave up \(\implies\) Minimum).
Therefore, \( p = 2 \).

Step 4: Calculate the y-coordinate (q)
Substitute \( x = 2 \) into the original function:
\( q = f(2) = \frac{4}{3}(2)^3 – 16(2) \)
\( q = \frac{4}{3}(8) – 32 \)
\( q = \frac{32}{3} – \frac{96}{3} \)
\( q = -\frac{64}{3} \)

Final Answer:
\( \boxed{p = 2} \)
\( \boxed{q = -\dfrac{64}{3}} \)