IB Mathematics SL 5.4 Tangents and normal AI SL Paper 1- Exam Style Questions- New Syllabus
Question
(i) the local maximum point.
(ii) the local minimum point.
(i) the gradient of this tangent.
(ii) the equation of this tangent, giving your answer in the form \( y = mx + c \).
Most-appropriate topic codes (IB Mathematics: Applications and Interpretation 2025):
• SL 5.2: Increasing and decreasing functions — part (b)
• SL 5.3: Derivative of power functions — part (c)
• SL 5.4: Equations of tangents at a given point — part (c)(ii)
• SL 5.6: Local maximum and minimum points where \( f'(x) = 0 \) — parts (a)(i), (a)(ii)
▶️ Answer/Explanation
(a)(i)
The local maximum occurs where \( f'(x) = 0 \) and the second derivative is negative. From the graph or using calculus: \( f'(x) = 3x^2 + 4x – 4 \). Solving \( f'(x) = 0 \) gives critical points. The local maximum is at \( x = -2 \).
\( \boxed{-2} \).
(a)(ii)
The local minimum is at \( x = \frac{2}{3} \) (or approximately 0.667).
\( \boxed{\frac{2}{3}} \) (or 0.667).
(b)
The function is decreasing where \( f'(x) < 0 \), which is between the two turning points.
\( \boxed{-2 < x < \frac{2}{3}} \).
(c)(i)
The tangent at \( (1, -3) \) is parallel to \( y = 3x + 5 \), so its gradient is the same as the slope of that line, which is 3.
\( \boxed{3} \).
(c)(ii)
Using point-gradient form: \( y – y_1 = m(x – x_1) \). With \( m = 3 \) and point \( (1, -3) \):
\( y + 3 = 3(x – 1) \)
\( y = 3x – 6 \).
\( \boxed{y = 3x – 6} \) or equivalent.