Question

Write the domain using interval notation.

Answer 1

`(f \circ g)(\text{x}) = \frac{13}{13-\text{x}}`

Domain:
`(-\infty,0) \cup (0,13) \cup (13,\infty)`

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Step-by-step explanation:

Let's find the function composition.

The notation
`(f \circ g)(\text{x})` is the same as
`f(g(\text{x}))`

`f(\text{x}) = \frac{\text{x}}{\text{x}-1}\\\\\\f(g(\text{x})) = \frac{g(\text{x})}{g(\text{x})-1}\\\\\\f(g(\text{x})) = g(\text{x}) / \Big( g(\text{x}) - 1\Big)\\\\\\`

Then,

`f(g(\text{x})) = \frac{13}{\text{x}} / \left(\frac{13}{\text{x}}-1}\right)\\\\\\f(g(\text{x})) = \frac{13}{\text{x}} / \left(\frac{13}{\text{x}}-\frac{\text{x}}{\text{x}}\right)\\\\\\f(g(\text{x})) = \frac{13}{\text{x}} / \frac{13-\text{x}}{\text{x}}\\\\\\f(g(\text{x})) = \frac{13}{\text{x}} * \frac{\text{x}}{13-\text{x}}\\\\\\f(g(\text{x})) = \frac{13}{13-\text{x}}\\\\\\`

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Now let's find the domain.

If we plugged x = 0 into g(x), then we get a division by zero error.

This means we must exclude this value from the domain.

For similar reasoning, we must exclude x = 13 because we get a division by zero error in
`f(g(\text{x})) = \frac{13}{13-\text{x}}`

We could have any other real number to be plugged in for x.

Here's what the domain looks like in interval notation.

`(-\infty,0) \cup (0,13) \cup (13,\infty)`

We effectively poke holes at 0 and 13 on the number line.