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Functions-Composition

by Gmat Bond » Wed Oct 07, 2015 1:38 am
Q:If g(x)=x^2-x+7, and g(f(x))=9x^2+9x+9, then f(x) is ? OA 3x+2.

How can we solve this question algebraically without plugging back the option into the original expression ? Request an algebraic solution. Please do not plug back the answer into the original expression. Thanks in advance !
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by Jim@StratusPrep » Wed Oct 07, 2015 7:00 am
My honest opinion is that you should not worry about the algebra for this, as it is not something that you will see on the GMAT.

If you are thinking of doing something like this on the GMAT (i.e. not plugging in answers) you are thinking about the exam the wrong way.
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by Max@Math Revolution » Thu Oct 08, 2015 12:59 am
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


Q:If g(x)=x^2-x+7, and g(f(x))=9x^2+9x+9, then f(x) is ?



Hi! Bond, this is my algebraic approach.

First of all g(x) and g(f(x)) is polynomials, we may assume that f(x) is also a polynomial with variable x.
Now since degree of g(x) is 2 and degree of g(f(x))=f^2 -f +7 is also 2, the degree of f(x) should be 1. That means f(x)= (a*x +b) ---> g(f(x))=(a*x +b)^2 -(a*x +b) +7. So the leading coefficient(the coefficient of highest degree variable) of (a*x +b)^2 -(a*x +b) +7 is a^2. It should be 9 ----> a^2 = 9 ---> a= 3 or -3.

case1) a=3, f(x)=3x+b.
g(f(x))= (3x+b)^2 -(3x+b) +7 = 9x^2 +(6b-3)x +(b^2-b+7)=9x^2+9x+9.
So 6b-3=9 ---> b=2 ---->2^2-2 +7 =9(satisfies the condition).
So f(x) = 3x+2

case2) a=-3, f(x)=-3x+b
Similarly with case1)
g(f(x))= (-3x+b)^2 -(-3x+b) +7 = 9x^2 +(3-6b)x +(b^2-b+7)=9x^2+9x+9.
So 3-6b=9 ---> b=-1 ---->(-1)^2-(-1) +7 =9(satisfies the condition).
So f(x) = -3x-1.

The answer is, therefore, 3x+2 or -3x-1.

Good luck to you Bond, GMAT BOND !!


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by Gmat Bond » Thu Oct 08, 2015 6:30 am
Max@Math Revolution wrote:Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


Q:If g(x)=x^2-x+7, and g(f(x))=9x^2+9x+9, then f(x) is ?



Hi! Bond, this is my algebraic approach.

First of all g(x) and g(f(x)) is polynomials, we may assume that f(x) is also a polynomial with variable x.
Now since degree of g(x) is 2 and degree of g(f(x))=f^2 -f +7 is also 2, the degree of f(x) should be 1. That means f(x)= (a*x +b) ---> g(f(x))=(a*x +b)^2 -(a*x +b) +7. So the leading coefficient(the coefficient of highest degree variable) of (a*x +b)^2 -(a*x +b) +7 is a^2. It should be 9 ----> a^2 = 9 ---> a= 3 or -3.

case1) a=3, f(x)=3x+b.
g(f(x))= (3x+b)^2 -(3x+b) +7 = 9x^2 +(6b-3)x +(b^2-b+7)=9x^2+9x+9.
So 6b-3=9 ---> b=2 ---->2^2-2 +7 =9(satisfies the condition).
So f(x) = 3x+2

case2) a=-3, f(x)=-3x+b
Similarly with case1)
g(f(x))= (-3x+b)^2 -(-3x+b) +7 = 9x^2 +(3-6b)x +(b^2-b+7)=9x^2+9x+9.
So 3-6b=9 ---> b=-1 ---->(-1)^2-(-1) +7 =9(satisfies the condition).
So f(x) = -3x-1.

The answer is, therefore, 3x+2 or -3x-1.

Good luck to you Bond, GMAT BOND !!


Math Revolution : Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World's First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Unlimited Access to over 120 free video lessons - try it yourself (https://www.mathrevolution.com/gmat/lesson)
See our Youtube demo (https://www.youtube.com/watch?v=R_Fki3_2vO8)
Hi! Max,

I have also devised an algebraic solution, but don't know whether my approach is correct. So advice me whether this approach is correct:-

g(x)=x^2-x+7 so g(f(x))=[f(x)]^2-f(x)+7 which is equal to 9x^2+9x+9

So we have: g(f(x))=[f(x)]^2-f(x)+7=9x^2+9x+9
or [f(x)]^2-f(x)=9x^2+9x+2
or f(x)[f(x)-1]=(3x+2)(3x+1)
Now, by comparison we can clearly see that f(x)=3x+2 and f(x)-1=3x+1.

You can see that we got the correct answer by simple algebraic manipulation. But I am not sure whether this method is correct. So I request your advice to guide me whether this method is correct and whether this method will always produce the correct answer. And whether we can always use this method safely for this type of questions? Please advice !
The name is Bond, GMAT BOND !!
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