Question

If `s(x)=x^2+1` and `t(x)=x-3`, does `s(t(x))=t(s(x))`?

Answer 1

No

Explanation:

Given
`color(white)("XXX")color(red)(s(x))=color(red)(x^2+1)`
and
`color(white)("XXX")color(blue)(t(x))=color(blue)(x-3)`

Then
`color(white)("XXX")s(color(blue)(t(x)))=(color(blue)(t(x)))^2+1`
`color(white)("XXXXXXX")=(color(blue)(x-3))^2+1`
`color(white)("XXXXXXX")=x^2-6x+8`
and
`color(white)("XXX")t(color(red)(s(x)))=color(red)(x^2+1)-3`
`color(white)("XXXXXXX")=x^2-2`

Clearly `s(t(x))!=t(s(x))`

Answer 2

No.
`s(t(x)) = x^2-6x+10`, while `t(s(x)) = x^2-2`

Explanation:

`s(t(x)) =(x-3)^2+1` (plug `t(x)` in the place of x in `s(x)`
`s(t(x)) = x^2-6x+9+1` (square `(x-3)`)
`s(t(x)) = x^2-6x+10` (simplify)

`t(s(x)) = (x^2+1)-3` (plug in `s(x)` in the place of x in `t(x)`)
`t(s(x)) = x^2-2` (simplify)