Answer:
Solution
verified
Verified by Toppr
f,g:R→R is defined as
f(x)=x+1
g(x)=2x−3
Now, (f+g)(x)=f(x)+g(x)=(x+1)+(2x−3)=3x−2
∴(f+g)(x)=3x−2
Now, (f−g)(x)=f(x)−g(x)=(x+1)−(2x−3)=x+1−2x+3=−x+4
∴(f−g)(x)=−x+4
(
g
f
)(x)=
g(x)
f(x)
,g(x)
=0,x∈R
∴(
g
f
)(x)=
2x−3
x+1
,2x−3
=0 or 2x
=3
∴(
g
f
)(x)=
2x−3
x+1
,x
=
2
3
