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

The function g is linear in two segments over its domain.

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

Hence, the range of g is:

$\text{Range of } g = \{ y \;|\; -9 \leq y \leq 4 \}.$

From the graph, we can see that at $x = 2$, the value of g is 0.

Thus:

$g(2) = 0.$

First, we find g(8) using the function g. The value at 8 on the graph is 4.

Now we find f(g(8)), which is f(4):

$f(4) = \frac{3 - 2(4)}{4 - 5} = \frac{3 - 8}{-1} = 5.$

So:

$fg(8) = 5.$

To sketch this graph, we take the piecewise definition of g:

1. For $-1 \leq x < 2$, g(x) is linear, decreasing from $(-1, -9)$ to $(2, 0)$, thus $|g(x)|$ will flip the negative part to positive.
2. From $(2, 0)$ to $(8, 4)$, it remains the same.

Thus, the graph of $y = |g(x)|$ will touch the x-axis at $(2, 0)$ and peak at $(8, 4)$.

To sketch the inverse, we reflect the graph of g over the line $y = x$:

1. Identify key points on g: (-1, -9), (2, 0), and (8, 4) become (-9, -1), (0, 2), and (4, 8) respectively.
2. Connect these plotted points in a linear manner.

This graph will start from $(-9, -1)$ to $(0, 2)$ and then from $(0, 2)$ to $(4, 8)$. It should have the corresponding shape to reflect the original function g.

The domain of the inverse function $g^{-1}$ corresponds to the range of g. Thus, the domain is:

$\text{Domain of } g^{-1} = \{ y \;|\; -9 \leq y \leq 4 \}.$