How do you find the domain of [fog](x) given f(x)=sqrt(x-2) and g(x)=1/(4x)?

1 Answer
Nov 25, 2016

The domain of (f@g)(x) is 0ltxlt1/8.

Explanation:

First, find (f@g)(x). Recall that (f@g)(x) is the same thing as f(g(x)), so take g(x)=1/(4x) and plug it into x in f(x)=sqrt(x-2).

So:

f(color(blue)x)=sqrt(color(blue)x-2)

f(color(blue)(g(x)))=sqrt(color(blue)(g(x))-2)=sqrt(1/(4x)-2)

Simplifying the fraction in the square root:

f(g(x))=sqrt((1-8x)/(4x))

We have a square root function. Note that when there is a square root, the contents of the square root have to be positive, or greater than 0. This means that our domain is restricted to:

(1-8x)/(4x)>0

There are multiple places to analyze here. The numerator is 0 when 1-8x=0 or x=1/8.

The denominator is 0 when 4x=0 at x=0. There is also a domain restriction at x=0 because this would make the denominator 0.

We need three intervals--one when x<0, another when 0ltxlt1/8, and the other when x>1/8.

Test the sign of (1-8x)/(4x) when x<0. Our example can be x=-1.

(1-8(-1))/(4(-1))=(1+9)/(-4)=-5/2

Since this is <0, this entire interval is excluded from the domain.

Testing the sign at 0ltxlt1/8, we can test x=1/16.

(1-8(1/16))/(4(1/16))=(1-1/2)/(1/4)=2

This is positive. Thus 0ltxlt1/8 is included in our domain.

Testing x>0 with x=1, we see that the sign of the portion inside the square root is

(1-8(1))/(4(1))=-7/4

So xgt0 is excluded from our domain as well.

The domain of (f@g)(x) is 0ltxlt1/8.