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

Question

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

1320 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-17T21:12:47

D)

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)

Answer 2 · 2018-03-18T09:18:17

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

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#