Question

If `f(x) = |x|` and `g(x) = (21)/(x^2+7)`, what is `(f@g)(4)`?

Answer 1

`(f@g)(4)=21/23`

Explanation:

Note that

`(f@g)(4)=f(g(4))`

So, we want to plug `g(4)` into `f(x)=absx`:

`f(g(4))=abs(g(4))`

`f(g(4))=abs(21/(4^2+7))`

`f(g(4))=21/23`