Question

If `f( x ) = \frac { 4} { x - 7)` and `g ( x ) = \frac { 5} { 7x }`, what is `(f*g)(x)`?

Answer 1

See the entire explanation below:

Explanation:

`(f * g)(x) = f(x) * g(x)`

Therefore:

`(f*g)(x) = 4/(x - 7) * 5/(7x)`

`(f*g)(x) = (4 * 5)/(7x(x - 7))`

`(f*g)(x) = 20/((7x * x) - (7x * 7))`

`(f*g)(x) = 20/(7x^2 - 49x)`

Answer 2

Depending upon what was meant:
`(f*g)(x)=(fxxg)(x)=20/(7x^2-49x)`
or
`(fcircg)(x)=f(g(x))=(28x)/(5-49x)`

Explanation:

The form of your question is confusing. I am not certain if you really meant `(f"*"g)(x)` (as you wrote), meaning `(fxxg)(x)` or if you meant the composite form `(fog)(x)`, meaning `f(g(x))`.

Answer 1: If you really meant `f` multiplied by `g`
Given
`color(white)("XXX")f(x)=4/(x-7)`
`color(white)("XXX")g(x)=5/(7x)`
then
`color(white)("XXX")(f xx g)(x)=4/(x-7)xx5/(7x)=20/(7x^2-49x)`

**Answer 2: If you intended `(fog)(x)`
Given
`color(white)("XXX")f(color(red)(x))=4/(color(red)(x)-7)`
`color(white)("XXX")color(blue)(g(x))=color(blue)(5/(7x))`
replacing `color(red)(x)` in the definition of `f(color(red)(x))`
with `color(blue)(g(x))`
`color(white)("XXX")(fog)(x) = f(color(blue)(g(x)))=4/(color(blue)(g(x))-7)`

`color(white)("XXXXXXXXXXXXXX")=4/(color(blue)(5/(7x))-7)`

`color(white)("XXXXXXXXXXXXXX")=4/((5-49x)/(7x))`

`color(white)("XXXXXXXXXXXXXX")=4xx(7x)/(5-49x)`

`color(white)("XXXXXXXXXXXXXX")=(28x)/(5-49x)`