Question

For `f(x) = \sqrt x` and `g(x) = x- 1`, what is `(f * g)(x); (g * f)(x); (f * g) (2) and (g * f)(2)`?

Answer 1

`sqrt(x-1)` ; `sqrt(x)-1`; `1` ; `0.41`

Explanation:

`(f*g)(x)` can be rewritten as `f(g(x))`
This means that for an x-value, you input that into `g(x)`, and then input your solution for that into `f(x)`.
`f(g(2))=sqrt(2-1)`
`sqrt(1)=1`
`(g*f)(x)` can be rewritten as `g(f(x)).` This time we start with `f(x)` and plug that into `g(x).`
`g(f(2))=(sqrt(2))-1` (Its good to memorize that `sqrt(2)~~1.41`)
`1.41-1`
`2`