IB Mathematics SL 2.5 Composite functions fog AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)
(i) To find \( f(4) \), locate \( x = 4 \) on the x-axis and trace vertically to the curve. The graph shows \( y \approx 1 \).
Answer: \( f(4) = 1 \)
(ii) To find \( f \circ f(4) \), first use \( f(4) = 1 \). Then, find \( f(1) \) by locating \( x = 1 \) on the x-axis and tracing vertically. The graph shows \( y \approx 3 \).
Answer: \( f \circ f(4) = f(1) = 3 \)
(iii) To find \( f^{-1}(3) \), find the \( x \)-value where \( f(x) = 3 \). Locate \( y = 3 \) on the y-axis and trace horizontally to the curve. The graph shows \( x \approx 1 \).
Answer: \( f^{-1}(3) = 1 \)
Part (b)
As per the markscheme, sketch the graph of \( y = f^{-1}(x) \):
– The graph is a concave up curve.
– It has a \( y \)-intercept at \( (0, 10) \).
– It has an \( x \)-intercept at \( (5, 0) \).
– The curve passes through either \( (2, 2) \) or both \( (1, 4) \) and \( (3, 1) \).
▶️ Answer/Explanation
Part (a)
(i) To find \( f(2) \), locate \( x = 2 \) on the x-axis and trace vertically to the curve. The graph shows \( y = 6 \).
Answer: \( f(2) = 6 \)
(ii) To find \( (f \circ f)(2) \), first use \( f(2) = 6 \). Then, find \( f(6) \) by locating \( x = 6 \) on the x-axis and tracing vertically. The graph shows \( y = -2 \).
Answer: \( (f \circ f)(2) = f(6) = -2 \)
Part (b)
Sketch the graph of \( g(x) = \frac{1}{2} f(x) + 1 \) for \( -4 \leq x \leq 6 \):
– The function \( g(x) \) transforms \( f(x) \) by scaling the y-values by \( \frac{1}{2} \) (vertical compression) and shifting upward by 1 unit.
– For key points on \( f(x) \), apply the transformation:
- At \( x = -4 \), \( f(-4) \approx 2 \), so \( g(-4) = \frac{1}{2} \cdot 2 + 1 = 2 \).
- At \( x = 2 \), \( f(2) = 6 \), so \( g(2) = \frac{1}{2} \cdot 6 + 1 = 4 \).
- At \( x = 6 \), \( f(6) = -2 \), so \( g(6) = \frac{1}{2} \cdot (-2) + 1 = 0 \).
– The shape of \( g(x) \) resembles \( f(x) \) but is vertically compressed and shifted up.
