Respuesta :
For this case we have the following equations:
[tex]f (x) = \frac {1} {x}\\g (x) = x ^ 2-3x[/tex]
We must find [tex](f_ {o} g) (x):[/tex]
By definition of composition of functions we have to:
[tex](f_ {o} g) (x) = f (g (x))[/tex]
So:
[tex](f_ {o} g) (x) = \frac {1} {x ^ 2-3x}[/tex]
We must find the domain of f (g (x)). The domain will be given by the values for which the function is defined, that is, when the denominator is nonzero.
[tex]x ^ 2-3x = 0\\x (x-3) = 0[/tex]
So, the roots are:
[tex]x_ {1} = 0\\x_ {2} = 3[/tex]
The domain is given by all real numbers except 0 and 3.
Answer:
x other than 0 and 3
ANSWER
0 and 3
EXPLANATION
The given functions are
[tex]f(x) = \frac{1}{x} [/tex]
and
[tex]g(x) = {x}^{2} - 3x[/tex]
[tex]( f \circ g)(x) = f(g(x))[/tex]
[tex]( f \circ g)(x) = f( {x}^{2} -x )[/tex]
[tex]( f \circ g)(x) = \frac{1}{ {x}^{2} - 3x} [/tex]
Factor the numerator:
[tex]( f \circ g)(x) = \frac{1}{ x(x - 3)} [/tex]
The function will be undefined if the denominator is zero.
[tex]x(x - 3) \ne0[/tex]
[tex]x \ne0 \: and \: x \ne3[/tex]
Therefore 0 and not in the domain of the composed function.