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Use the given functions to find f (g(x)), and give the restrictions on x.

F(x)= 1/x-3 and g(x) = 3/x +3


F(g(x)) = ?


Where x(doesn’t = ?

Answer :

Answer:

[tex]f(g(x)) = \frac{10x-9}{3 + 3x}[/tex]

Step-by-step explanation:

Given

[tex]f(x) = \frac{1}{x} - 3[/tex]

[tex]g(x) = \frac{3}{x} + 3[/tex]

Required

Find f(g(x))

If: [tex]f(x) = \frac{1}{x} - 3[/tex]

Then:

[tex]f(g(x)) = \frac{1}{3/x + 3} -3[/tex]

Solve the denominator (take LCM)

[tex]f(g(x)) = \frac{1}{\frac{3+3x}{x}} -3[/tex]

[tex]f(g(x)) = \frac{x}{3 + 3x} - 3[/tex]

It can be solved further as:

[tex]f(g(x)) = \frac{x-3(3 + 3x)}{3 + 3x}[/tex]

[tex]f(g(x)) = \frac{x-9 + 9x}{3 + 3x}[/tex]

[tex]f(g(x)) = \frac{10x-9}{3 + 3x}[/tex]

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Rewritten by : Barada