Question

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

Answer 1

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`.