Question

Find the indicated values where g(t)=t2-t And f(x) =1+x
g(f(0)) + f(g(0))

Answer 1

Answer:

$\boxed{g(f(0))+ f(g(0))=1}$

Step-by-step explanation:

This is a problem of composition functions.

$The \ \mathbf{composition} \ of \ the \ function \ f \ with \ the \ function \ g \ is:\\ \\ (f \circ g)(x)=f(g(x)) \\ \\ The \ domain \ of \ (f \circ g) \ is \ the \ set \ of \ all \ x \ in \ the \ domain \ of \ g \\ such \ that \ g(x) \ is \ in \ the \ domain \ of \ f$

On the other hand:

$(g \circ f)(x)=g(f(x))$

So let's find each value as follows:

$f(0)=1+0=1 \\ \\ g(0)=0^2-0=0 \\ \\ Therefore: \\ \\g(f(0))=g(1)=1^2-1=0 \\ \\ f(g(0))=f(0)=1 \\ \\ Finally: \\ \\ g(f(0))+ f(g(0))=0+1 \\ \\ \therefore \boxed{g(f(0))+ f(g(0))=1}$

Answer 2

Answer:

Choice A is correct

Step-by-step explanation:

The composite function g(f(x)) will be given by (1+x)^2 -(1+x). Substituting x with zero yields g(f(0)) =0. On the other hand, the composite function f(g(t)) will be given by 1+t^2-t. Substituting t with zero yields f(g(0)) =1. The sum is thus equal to 1