IBDP Maths SL 5.6 Derivatives of xn , sinx , cosx , tanx AA HL Paper 2- Exam Style Questions- New Syllabus
Question
(a) Determine the set of values of \( x \) for which the function \( f \) is decreasing.
(b) Find the interval of \( x \) for which the graph of \( f \) is concave-up.
Syllabus Topic Codes (IB Mathematics AA HL):
• SL 5.6: Derivative of \( e^x \) — parts (a), (b)
• SL 5.7: The second derivative— part (b)
• SL 5.8: Concavity — part (b)
▶️ Answer/Explanation
(a)
\( f \) is decreasing when \( f'(x) < 0 \):
\( 4 + 2x – 3e^x < 0 \).
Solving \( 4 + 2x – 3e^x = 0 \) gives critical values
\( x \approx -1.73554 \) and \( x \approx 0.517999 \).
Testing intervals shows \( f'(x) < 0 \) for
\( x < -1.73554 \) and \( x > 0.517999 \).
\( x \leq -1.74 \) and \( x \geq 0.518 \) (or strict inequalities).
(b)
Concave-up when \( f”(x) > 0 \).
\( f”(x) = 2 – 3e^x \).
Solve \( 2 – 3e^x = 0 \): \( e^x = \frac{2}{3} \), so \( x = \ln\!\left(\frac{2}{3}\right) \approx -0.405465 \).
\( f”(x) > 0 \) when \( 2 – 3e^x > 0 \implies e^x < \frac{2}{3} \implies x < \ln\!\left(\frac{2}{3}\right) \).
\( x < \ln\!\left(\frac{2}{3}\right) \) or \( x < -0.405 \).