# Composite function problem.

1. Apr 1, 2012

### valmancer

Given f(x)=ax+b, and f(3)(x)=64x+21, find the values of the constants a and b.

(note: f(3)(x) means fff(x))

To me this seems like I have to use two equations to find the value of three variables, since when I have found a and b, I should be able to get the value of x. Even though it should be impossible to find the values of three variables (I don't know what the difference between an unknown constant and a variable is) with two equations I went ahead and started simplifying the equation hoping it to magically solve itself:

f(3)(x)=a(a(ax+b)+b)+b=a(a^2x+ab+b)+b

=>a^3x+a^2b+ab+b=64x+21

=>b(a^2+a+1)=64x+21-a^3x

=>b=(64x+21-a^3x)/(a^2+a+1)

And this is where I understood that the way I was going about solving the thing won't work (I need to find exact values, which are not dependent on any variables). I'm out of ideas for how to try and solve the question.

2. Apr 1, 2012

### Solaris69

After a^3x+a^2b+ab+b=64x+21 you should have two equations:

a^3 = 64
and a^2b+ab+b = 21

---> a = 4
and 4^2b + 4b +b = 21
16b + 4b +b = 21
21 b = 21
b = 21

3. Apr 1, 2012

### Bearded Man

Do you know what makes two polynomials equal? Two polynomials are equal if and only if their coefficients with respect to the variables are equal. For example, if
$$ax^2 + b = 16x^2 + 4$$
then $$a=16$$ and $$b = 4$$

4. Apr 1, 2012

### valmancer

Thanks a lot; I don't understand how I didn't notice that.

This is how I got the answer in case anyone's interested:

a^3x+(a^2b+ab+b)=64x+21

Therefore a^3=64 => a=4

a^2b+ab+b=21
=> b(4^2+4+1)=21 => b=21/21=1

Thank you