Find fog and gof, if

f(x) = x + 1, g(x) = 2x + 3

f(x) = x + 1 and g(x) = 2x + 3

Range of f = R ⊂ Domain of g = R ⇒ gof exists

Range of g= R ⊂ Domain of f ⇒ fog exists

Now,

fog(x) = f(g(x) = f(2x + 3) = (2x + 3) + 1= 2x + 4 and

gof(x)= g(f(x))= g(x + 1) = 2(x + 1) + 3 = 2x + 5

So, fog(x) = 2x + 4 and gof(x) = 2x + 5