# Let f(x)= 12-4x and g(x)= 1/x.......what is the domain of g(f (x))

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Given the functions:

f(x) = 12 - 4x

g(x) = 1/x

We need to find the domain of g(f(x)).

First we need to determine g(f(x))

g(f(x)) = g( 12 - 4x).

Now we will substitute ( 12 - 4x ) in g(x).

==> g(f(x)) = 1/ ( 12-4x)

Let us simplify the denominator:

==> g(f(x)) = 1/4(3-x)

Now the domain are x values such that g(f(x)) is defined.

g(f(x)) is **not** defined when the denominator = 0.

Then we will find the values where g(f(x)) is not defined.

==> 4(3-x) = 0

==> 3 - x= 0

==> x= 3

Then the only value of x where the function is not defined is x= 3.

Then x is not in the domain of f(g(x)).

**Then the domain is x = R - { 3} **

We'll determine the result of the composition of the functions:

(gof)(x) = g(f(x))

That means that in the expression of g(x), we'll substitute x by f(x):

g(f(x)) = 1/f(x)

We'll re-write the result putting the expression of f(x) at denominator of the composed function:

1/f(x) = 1/(12-4x)

Now, we'll impose the constraint that for the ratio to exist, the denominator has to be different from zero.

More accurate, we'll compute the value of x that makes the denominator to cancel and we'll reject them form the domain of the function g(f(x)).

12 - 4x = 0

We'll subtract 12 both sides:

-4x = -12

x = -12/-4

x = 3

**So, the domain of definition of the composed function is the real set R, without the value of 3.**

**x belongs to the interval (-infinite ; +infinite) - {3}**

f(x) = 12-4x and g(x) = 1/x

To find g(f(x)) .

g(x) = 1/x.

We substitute 12-x in place of x in 1/x to obtain g(f(x)).

Therefore g(f(x)) = 1/(12-x)..