IBDP Maths SL 2.2 Function and their domain range graph AA HL Paper 1- Exam Style Questions- New Syllabus
Question
(b) Write down the domain of \( f^{-1} \), the inverse function of \( f \).
(c) Determine the value of \( x \) for which \( f^{-1}(2x – 7) = -3 \).
Most-appropriate topic codes (IB Mathematics: Analysis and Approaches HL 2025):
▶️ Answer/Explanation
(a)
From the graph, when \( x = -3 \), the corresponding \( y \)-value is \( -1 \). Therefore, \( f(-3) = -1 \).
\(\boxed{-1}\)
(b)
The domain of the inverse function \( f^{-1} \) is equal to the range of the original function \( f \). From the graph, the range of \( f \) is from \( y = -1 \) to \( y = 5 \), inclusive. Therefore, the domain of \( f^{-1} \) is \( -1 \le x \le 5 \). In interval notation: \( [-1, 5] \).
\(\boxed{-1 \le x \le 5}\) or \(\boxed{[-1, 5]}\)
(c)
We are given \( f^{-1}(2x – 7) = -3 \).
Apply the function \( f \) to both sides:
\( f(f^{-1}(2x – 7)) = f(-3) \)
Since \( f(f^{-1}(u)) = u \) for \( u \) in the domain of \( f^{-1} \), this simplifies to:
\( 2x – 7 = f(-3) \)
From part (a), \( f(-3) = -1 \), so:
\( 2x – 7 = -1 \)
\( 2x = 6 \)
\( x = 3 \)
\(\boxed{3}\)