The function f is defined by $$f: x \mapsto \frac{3 - 2x}{x - 5}, \; x \in \mathbb{R}, \; x \neq 5$$ (a) Find $f^{-1}(x)$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 4

The function g is linear within the intervals defined. Observing the line segments from the graph:

• For the segment from $(-1, -9)$ to $(2, 0)$, the range covers values from $-9$ to $0$.
• For the segment from $(2, 0)$ to $(8, 4)$, the range extends from $0$ to $4$.

Thus, the overall range of g is: $-9 \leq g(x) \leq 4$

From the graph of g, we see that: $g(2) = 0$

First, we determine g(8) from the graph, which shows: $g(8) = 4$.

Next, we substitute into f: $f(g(8)) = f(4) = \frac{3 - 2(4)}{4 - 5} = \frac{3 - 8}{-1} = \frac{-5}{-1} = 5$

To sketch $y = |g(x)|$, we take the absolute values of the y-coordinates of g. This will reflect any negative part of the function above the x-axis.

• The points $(-1, -9)$ will become $(-1, 9)$.
• The segment from $(-1, 9)$ to $(2, 0)$ remains as is.
• The range from $(2, 0)$ to $(8, 4)$ remains unchanged.

For coordinates where it meets the axes:

• The function cuts the x-axis at $(2, 0)$ and $(8, 4)$.
• The point $(0, 6)$ results from the vertical intercept of the graph after reflecting the negative values.

Overall, this transforms negative values to positive on the graph with key points marked appropriately.

To sketch $y = g^{-1}(x)$, we invert the axes of g. The important points are:

1. $(0, 2)$ and $(2, 0)$ must be marked, indicating where the function switches.
2. The vertical intercept can be traced at $(-6, 0)$ from the reflection.

Overall, we include coordinates at points $(2, 0)$ and $(-6, 0)$, while ensuring the shape of the graph retains its linearity.

The domain of $g^{-1}$ corresponds to the range of the original function g. Based on the range information derived earlier, the domain of the inverse function is: $-9 \leq x \leq 4$