Domain of a functionLet f(x)= 12-4x and g(x)= 1/x...What is the domain of g(f (x))?

3 Answers

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to find the domain of the new function `g(f(x))` , hence, you need to find the equation of this function replacing the equation of `f(x)` for x in equation of `g(x),` such that:

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

You need to determine the restrictions on the domain of the composition, hence, you need to determine the elements that create zero denominator, thus, you need to solve algebraically the following equation, such that:

`12 - 4x = 0 => 12 = 4x => x = 12/4 => x = 3`

Hence, evaluating the domain of the composition `g(f(x))` yields R with exclusion of 3.

tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

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

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

= 1/(12 - 4x)

= 1/(4*(3 - x))

The domain of a function f(x) is the set of values that x can take for which f(x) is real and defined.

For g(f(x)) = 1/(4*(3 - x)), the expression is not defined only when x = 3 as that makes the denominator equal to 0.

The domain of g(f(x)) is R - {3}

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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}