If f(x)=x+1 and (f@g)(x)=x^2+3x+2, which of the following is g(x)?

A) x+2
B) x^2+3x
C)x^2+3x-1
D)x^2+3x+1
E)(x+1)^2(x+2)

2 Answers
Mar 17, 2018

D)

Explanation:

Proposing

g(x) = a x^2+b x+c

we have

f(g(x)) = a x^2+b x+c+1 = x^2+3x+2

now comparing coefficients

{(c=1),(b=3),(a=1):}

hence

g(x) = x^2+3x+1

so the answer is D)

Mar 18, 2018

g(x) = x^2+3x+1

Explanation:

Since the function f(x) is one-to-one and onto, it is invertible. In fact, it is easy to see that the inverse function is

f^-1(x) = x-1

Quick check : (f^-1 circ f)(x) = f^-1(f(x)) = f^-1(x+1)=x

Thus

g(x) = ((f^-1 circ f)circ g)(x) = (f^-1 circ(f circ g))(x)
qquad = f^-1((f circ g)(x)) = f^-1(x^2+3x+2) = (x^2+3x+2)-1

So g(x) = x^2+3x+1