Question

If f(x)=1/x and g(x)=x^2-4x. What two numbers are not in the domain of f o g?

Answer 1

The composed fraction
`f \circ g` means that the input of
`f(x)` is the ouput of
`g(x)`.

So, the chain is

`x \mapsto g(x) \mapsto f(g(x)) = f\circ g = (1)/(g(x)) = (1)/(x^2-4x)`

The domain of
`f(x)` is composed by all inputs which are not zero, otherwise we would have a zero denominator.

So, since
`f(x)` doesn't accept 0 as an input, and we want to feed it with
`g(x)`, we conclude that
`g(x)` can't be zero.

So, we have

`f \circ g = f(g(x)) = f(x^2-4x) = (1)/(x^2-4x) \implies x^2-4x \\eq 0`

Since

`x^2-4x = x(x-4)`

we have

`x^2-4x = 0 \iff x(x-4) = 0 \iff x=0 \lor x=4`

Since these two points cases
`g(x)` to vanish, they can't be accepted as inputs by
`f(x)`.