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

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#.