IB Mathematics SL 2.2 Function and their domain range graph AA SL Paper 1- Exam Style Questions- New Syllabus

(a) Find \( f(0) \):
Substitute \( x = 0 \) into the function expression:
\( f(0) = \frac{3(0) – 2}{2(0) + 1} = \frac{-2}{1} = -2 \).
Answer: \( \boxed{-2} \)

(b) Equation of the horizontal asymptote:
For a rational function \( f(x) = \frac{ax + b}{cx + d} \), the horizontal asymptote is given by \( y = \frac{a}{c} \)[cite: 1007].
Here, \( a = 3 \) and \( c = 2 \).
Answer: \( \boxed{y = \frac{3}{2}} \)

(c) Find the range of \( g \):
First, consider the behavior of \( f(x) \) for \( x \geq 0 \).
The function starts at \( f(0) = -2 \) and increases towards the horizontal asymptote \( y = \frac{3}{2} \).
Thus, for \( x \geq 0 \), the range of \( f \) is \( -2 \leq f(x) < \frac{3}{2} \)[cite: 952].

The function \( g(x) = -f(x) \) represents a reflection in the \( x \)-axis[cite: 1033].
Reflecting the range of \( f \) across the \( x \)-axis:
Lower bound: \( \frac{3}{2} \times (-1) = -\frac{3}{2} \) (non-inclusive).
Upper bound: \( -2 \times (-1) = 2 \) (inclusive).
Answer: \( \boxed{-\frac{3}{2} < g(x) \leq 2} \)