IB Mathematics SL 5.4 Tangents and normals AA SL Paper 1- Exam Style Questions- New Syllabus
Question
Let the function \( f \) be defined by \( f(x) = x^3 + 5x^2 – 8 \), for \( x \in \mathbb{R} \).
(a) Find \( f'(1) \).
(b) Find the equation of the tangent to the curve \( y = f(x) \) at the point where \( x = 1 \).
Most-appropriate topic codes (Mathematics: analysis and approaches guide):
• SL 5.3: Derivative of \( f(x) = ax^n + bx^{n-1} + \dots \) — Part a
• SL 5.4: Tangents and normals at a given point, and their equations — Part b
• SL 5.1: Derivative interpreted as gradient function and as rate of change — Part a, b
• SL 5.4: Tangents and normals at a given point, and their equations — Part b
• SL 5.1: Derivative interpreted as gradient function and as rate of change — Part a, b
▶️ Answer/Explanation
(a)
First, differentiate \( f(x) \) using the power rule:
\( f'(x) = 3x^2 + 10x \)
Then evaluate at \( x = 1 \):
\( f'(1) = 3(1)^2 + 10(1) = 3 + 10 = 13 \)
\( \boxed{13} \)
(b)
The tangent line passes through the point \( (1, f(1)) \). First, calculate the y-coordinate \( f(1) \):
\( f(1) = 1^3 + 5(1)^2 – 8 = 1 + 5 – 8 = -2 \)
The gradient of the tangent at \( x = 1 \) is \( f'(1) = 13 \).
Using the point-slope form \( y – y_1 = m(x – x_1) \):
\( y – (-2) = 13(x – 1) \)
\( y + 2 = 13x – 13 \)
\( y = 13x – 15 \)
\( \boxed{y = 13x – 15} \)