The function \( g(x) = 3 – \frac{5}{x + 4} \) is a rational function. To determine the range, we will analyze its behavior by considering the vertical and horizontal asymptotes as well as the function’s general behavior.

Observation 1: Identifying the Vertical Asymptote

The vertical asymptote occurs where the denominator equals zero, which makes the function undefined. Setting the denominator equal to zero, we solve: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] Therefore, there is a vertical asymptote at \( x = -4 \). As \( x \) approaches this value, the behavior of \( g(x) \) differs depending on the direction from which \( x \) approaches.

Observation 2: Behavior Near the Vertical Asymptote

– As \( x \to -4^+ \) (approaching from the right), the denominator \( (x + 4) \) becomes a small positive value, causing \( \frac{5}{x + 4} \) to grow large. Thus, \( g(x) \to -\infty \). – As \( x \to -4^- \) (approaching from the left), the denominator \( (x + 4) \) becomes a small negative value, causing \( \frac{5}{x + 4} \) to grow large negatively. Thus, \( g(x) \to +\infty \). Hence, the function experiences a discontinuity at \( x = -4 \), with \( g(x) \) jumping from \( +\infty \) to \( -\infty \) at this point.

Observation 3: Behavior as \( x \to \pm \infty \)

As \( x \to \pm \infty \), the term \( \frac{5}{x + 4} \) approaches zero because the denominator increases without bound. Therefore, the function \( g(x) \) approaches the horizontal asymptote: \[ g(x) \to 3 \] This means that as \( x \) becomes very large or very small, \( g(x) \) approaches 3, but it never actually reaches this value.

Observation 4: Determining the Range of \( g(x) \)

Given the vertical asymptote at \( x = -4 \) and the horizontal asymptote at \( y = 3 \), the function can take all values except 3. The function jumps from \( +\infty \) to \( -\infty \) at \( x = -4 \), and as \( x \) moves away from this point, it approaches 3 without reaching it. Therefore, the range of \( g(x) \) is: \[ g(x) \in (-\infty, 3) \cup (3, \infty) \]

1 mark for correctly identifying the vertical asymptote at \( x = -4 \).

1 mark for accurately explaining the behavior of \( g(x) \) as \( x \to -4^\pm \) and as \( x \to \pm \infty \).

1 mark for correctly concluding the range of the function as \( (-\infty, 3) \cup (3, \infty) \).

To find \( g^{-1}(1) \), we need to solve the equation \( g(x) = 1 \) for \( x \). Start by setting \( g(x) = 1 \): \[ 1 = 3 – \frac{5}{x + 4} \] Rearrange this equation to isolate the term involving \( x \): \[ \frac{5}{x + 4} = 2 \] Next, multiply both sides by \( x + 4 \) to eliminate the fraction: \[ 5 = 2(x + 4) \] Simplify the equation: \[ 5 = 2x + 8 \] Subtract 8 from both sides: \[ -3 = 2x \] Finally, divide both sides by 2 to solve for \( x \): \[ x = -1.5 \] Thus, the value of \( g^{-1}(1) \) is \( -1.5 \).

1 mark for correctly isolating \( x \) and solving the equation.

1 mark for providing the correct final answer \( g^{-1}(1) = -1.5 \).