Home / IB Mathematics SL 2.5 Composite functions fog AA SL Paper 1- Exam Style Questions

IB Mathematics SL 2.5 Composite functions fog AA SL Paper 1- Exam Style Questions

IB Mathematics SL 2.5 Composite functions fog AA SL Paper 1- Exam Style Questions- New Syllabus

Question

The graph of \( y = f(x) \) for \( 0 \leq x \leq 10 \) is shown below. The graph intercepts the axes at \( (10, 0) \) and \( (0, 5) \).

Graph of y = f(x)

Part (a):
Write down the value of
(i) \( f(4) \)
(ii) \( f \circ f(4) \)
(iii) \( f^{-1}(3) \)

Part (b):
On the axes above, sketch the graph of \( y = f^{-1}(x) \). Show clearly where the graph intercepts the axes.

▶️ Answer/Explanation
Detailed Solutions

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 \( (3, 1) \).

Graph of f(x) and f^-1(x)

Question

The graph of \( y = f(x) \) for \( -4 \leq x \leq 6 \) is shown below.

Graph of y = f(x)

Part (a):
Write down the value of
(i) \( f(2) \)
(ii) \( (f \circ f)(2) \)

Part (b):
Let \( g(x) = \frac{1}{2} f(x) + 1 \) for \( -4 \leq x \leq 6 \). On the axes above, sketch the graph of \( g \).

▶️ Answer/Explanation
Detailed Solutions

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.

Graph of g(x)

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