Question 1199428: Find f(x) if f(g(x))= (f(x))(x+1)(x+2)-1 and g(x) = x^3 +2x^2 -x -2
Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website! .
Find f(x) if f(g(x))= (f(x))(x+1)(x+2)-1 and g(x) = x^3 +2x^2 -x -2
~~~~~~~~~~~~~~
The problem's formulation looks strange and, probably (from my point of view), is incomplete.
To be complete, the problem must say that f(x) is a polynomial function.
But I will assume it " by default " or " from the context ".
Then from the degree analysis, f(x) must be a linear function (a binomial)
f(x) = ax + b,
and my task is to find coefficients "a" and "b".
For it, it is enough to know f(x) in two points.
I will choose the points x= -1 and x= 0, and will coplete the solution step by step.
(a) if x = -1, then g(-1) = (-1)^3 + 2*(-1)^2 - (-2) - 2 = -1 + 2 + 1 - 2 = 0.
Therefore, the given equality f(g(x))= (f(x))(x+1)(x+2)-1 takes the form
f(0) = (f(-1))*(-1 +1 )*(-1 + 2) - 1 = (f(-1))*0*1 - 1 = -1. (*)
(b) if x = 0, then g(0) = 0^3 + 2*0^2 - 0 - 2 = -2.
Therefore, the given equality f(g(x))= (f(x))(x+1)(x+2)-1 takes the form
f(-2) = (f(0))*(0 +1 )*(0 + 2) - 1 = (f(0))*1*2 - 1 = 2*f(0) - 1.
Substitute here f(0) = -1 from (*), and you will get
f(-2) = 2*(-1) - 1 = -2 - 1 = -3.
(c) Thus we have these two equations to find coefficients "a" and "b" in f(x) = ax + b
f(0) = a*0 + b = -1, (1)
f(-2) = a*(-2) + b = -3. (2)
Equation (1) gives b = -1.
Then equation (2) gives
-2a - 1 = -3 ---> -2a = -3 + 1 = -2 ---> a = 1.
Thus f(x) = x-1. ANSWER
CHECK. Then left side f(g(x)) is (x^3 +2x^2 -x -2) - 1 = x^3 + 2x^2 - x -3.
Right side is (x-1)*(x+1)*(x+2) - 1 = (x-1)*(x^2 + 3x + 2) - 1 =
= x^3 + 3x^2 + 2x - x^2 - 3x - 2 - 1 = x^3 + 2x^2 - x -3.
Both sides are the same (are identical).
Thus the answer is verified/confirmed.
Solved.
|
|
|
