IB Mathematics SL 5.2 Increasing and decreasing function AI SL Paper 1- Exam Style Questions- New Syllabus
Question
A function is defined by \( f(x) = 2x^{-2} + bx^{-1} + c \), where \( b, c \in \mathbb{R} \).
The gradient of the tangent to the curve \( y = f(x) \) is \( 0.208 \) at the point where \( x = 5 \).
(a) Find the value of the constant \( b \).
(b) By considering the sign of the derivative, justify why the function \( f(x) \) is increasing at \( x = 3.5 \).
Most appropriate topic codes (IB Mathematics: applications and interpretation):
• SL 5.3: Derivative of \( ax^n \) and related sums — part (a)
• SL 5.1: Derivative interpreted as a gradient function — part (a)
• SL 5.2: Identifying increasing and decreasing functions using the first derivative — part (b)
• SL 5.1: Derivative interpreted as a gradient function — part (a)
• SL 5.2: Identifying increasing and decreasing functions using the first derivative — part (b)
▶️ Answer/Explanation
(a)
Differentiate: \(f'(x) = -4x^{-3} – bx^{-2}\)
At \(x = 5\): \(f'(5) = -4(5)^{-3} – b(5)^{-2}\)
\(0.208 = -4 \times \frac{1}{125} – b \times \frac{1}{25}\)
\(0.208 = -0.032 – \frac{b}{25}\)
\(\frac{b}{25} = -0.032 – 0.208 = -0.24\)
\(b = -0.24 \times 25 = -6\)
\(\boxed{-6}\)
(b)
Using \(b = -6\): \(f'(x) = -4x^{-3} + 6x^{-2}\)
At \(x = 3.5\): \(f'(3.5) = -4(3.5)^{-3} + 6(3.5)^{-2}\)
\(= -4 \times 0.029155 + 6 \times 0.081633\)
\(= -0.11662 + 0.489798 = 0.373178\)
Since \(f'(3.5) \approx 0.373 > 0\), the gradient is positive, so \(f(x)\) is increasing at \(x = 3.5\).
\(f'(3.5) > 0\), therefore \(f(x)\) is increasing at \(x = 3.5\).