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.

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