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

$x \mapsto g(x) \mapsto f(g(x)) = f\circ g = \dfrac{1}{g(x)} = \dfrac{1}{x^2-4x}$

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

$f \circ g = f(g(x)) = f(x^2-4x) = \dfrac{1}{x^2-4x} \implies x^2-4x \neq 0$

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 $f(x)$.