Solving Composite Functions: Find g(3) with g(2)=37

[Saitama]

Homework Statement

Suppose f(x) and g(x) are non-zero polynomials with real coefficients, such that f(g(x))=f(x)⋅g(x). If g(2)=37, what is g(3)?

Homework Equations

The Attempt at a Solution

I have no idea where to begin this with.
At x=2,
f(g(2))=f(2).g(2)
f(37)=37f(2)

What should I do with this equation? I cannot even guess any function which would satisfy the above condition.

Any help is appreciated. Thanks!

 

[Simon Bridge]

You know they are polynomials ... what are the properties of polynomials?

Note: if $f=x^2+1$ and $g=x$, both polynomials,
then $f(g)=x^2+1$ and $f.g = x^3+x$ so $f(g)\neq f.g$ so these are not correct.

You have to try to figure out what sort of polynomial will fit in there.
Start by revising the properties of polynimials and then experiment with different short polynomials to get a feel for how they behave.

 

[Saitama]

You have to try to figure out what sort of polynomial will fit in there.
Start by revising the properties of polynimials and then experiment with different short polynomials to get a feel for how they behave.

Does that mean that there's no proper way of doing it and I have to guess the polynomials? D:

 

[Simon Bridge]

The proper way to do it is to use your understanding of the properties of polynomials.

Since you have yet to gain that understanding, you will have to make educated guesses ... refine your guesses as you go: do not make blind guesses - there are infinite possibilities to choose from. You are supposed to use your knowledge of polynomials to guide your choices.

You know things like that $g=\sum_{i=0}^n a_n x^n$ and $g(2)=\sum_{i=0}^n a_n 2^n = 37$ ... and you know how to do powers of 2.

Fortunately (you should have realized) f and g must be quite simple polynomials ... they won't be order n=1045 or anything like that. You know this because your instructor knows your competence level and has set the problem in keeping with it ;) Try guessing just simple powers of x ... what happens if f=x^2 and g=x^3 ? Do f and g have to be different?

eg. f=g=x^2, f.g=x^2, f(g)=x^4 ... so f(g)=fg ... looks good: unfortunately g(2)=4 so it's not right.
g = x^5 + x^2 + 1 works ... but so does: g=x+35
Get the idea?

You can also work with the relationships... eg.
if f(3)=2f(2) then what is f(x) likely to be?

That kind of thing. In your case you know that f(37)=37f(2).

There will also be clues in your coursework so far.

 

[ehild]

What can be the order of the polynomials?

ehild

 

[Saitama]

Sorry for being late.

The proper way to do it is to use your understanding of the properties of polynomials.

Since you have yet to gain that understanding, you will have to make educated guesses ... refine your guesses as you go: do not make blind guesses - there are infinite possibilities to choose from. You are supposed to use your knowledge of polynomials to guide your choices.

I don't think that I do have time to revise everything about polynomials as it is almost the end of session and I am done with my coursework. :)

eg. f=g=x^2, f.g=x^2, f(g)=x^4 ... so f(g)=fg ... looks good: unfortunately g(2)=4 so it's not right.
g = x^5 + x^2 + 1 works ... but so does: g=x+35
Get the idea?

You can also work with the relationships... eg.
if f(3)=2f(2) then what is f(x) likely to be?

That kind of thing. In your case you know that f(37)=37f(2).

There will also be clues in your coursework so far.

I too guessed those two same polynomials (and some more) for g(x) but not even one for f(x). I could only deduce that both f(x) and g(x) shouldn't be of same degree. For example, had they been both a linear polynomial, the LHS would be a linear one and the RHS would consist of a term containing x^2.

 

[ehild]

I could only deduce that both f(x) and g(x) shouldn't be of same degree.

Are you sure?

If f is of n degree and g is of m degree what is the degree of f(g(x))? and what is the degree of f(x)g(x)?

Starting with the highest degree term,

f(x) =a_nxⁿ+...
g(x)=b_mx_m+...

f(g(x))=a_n(b_mx^m+... )ⁿ+...=(a_nxⁿ+...)(b_mx^m+...)

What is the relation between n and m?

ehild

 

[eddybob123]

I don't think that there is only one solution. Since we only want g(x), and f(x) is allowed to be anything we want, then just guess a polynomial for g(x) such that g(2) = 37, find the corresponding f(x), then solve for g(3).

 

[Saitama]

Are you sure?

If f is of n degree and g is of m degree what is the degree of f(g(x))? and what is the degree of f(x)g(x)?

f(g(x)) has the degree mn and f(x).g(x) has the degree m+n.

What is the relation between n and m?

Is the relation $m=\frac{n}{n-1}$ ?

If this is the relation, I get only one solution, which is m=2 and n=2.

 

[Saitama]

Ooohoo, looks like I got it.

Using the above relation, m=2 and n=2, I assumed that f(x) is a quadratic of the form $ax^2+bx+c$.
Since, $f(37)=37 \cdot f(2)$
$$(37)^2a+37b+c=37(4a+2b+c)$$
Rearranging, I end up with
$$37 \times 33a-37b-36c=0$$
The possible values of $a,b,c$ can be $a=1, b=33$ and c=0.
Therefore, $f(x)=x^2+33x$
As $f(g(x))=f(x).g(x)$
$$(g(x))^2+33 \cdot g(x)=(x^2+33x) \cdot g(x)$$
The above relation is a quadratic in g(x) and from here I get two solutions.
$g(x)=0$ and $g(x)=x^2+33x-33$ but g(x) is a non-zero polynomial, hence $g(x)=0$ is not admissible. Both f(x) and g(x) satisfy the given conditions.
Therefore, g(3)=75.

Thanks a lot both of you.

 

[ehild]

Well, that is the solution, but the problem said that the coefficients are real. So the equation 37×33a−37b−36c=0 has infinite number of solutions. The coefficient of the quadratic term is arbitrary, take it a=1.

If f(x)=x²+bx+c and g(x) = Ax²+Bx+C

f(g(x))=(Ax²+Bx+C)²+b(Ax²+Bx+C)+c=(x²+bx+c)(Ax²+Bx+C)

You can rearrange the equation so as one side is quadratic.

b(Ax²+Bx+C)+c=(Ax²+Bx+C)(Ax²+Bx+C-x²-bx-c)

Expand, and compare the coefficients. Start with the highest degree terms. The coefficient of X⁴ is A(A-1)=0 so A=1. The coefficient of x³ is A(B-b), so B=b...

ehild

 

[Saitama]

... A(A-1)=0 so A=1.

Won't this equation give two values for A? And corresponding to each value, there will be a different value for B?

 

[ehild]

A can not be zero as it would make g(x) first order.

ehild

 

[vio]

?

frm which book did you get dis qstn??

 

[Saitama]

Sorry for the late reply.

A can not be zero as it would make g(x) first order.

ehild

Comparing coefficient of $x^2$,
$$Ab=A(C-c)+B(B-b)+C(A-1)$$
But A=1 and B=b,
$$b=C-c$$

Comparing coefficient of $x$,
$$Bb=B(C-c)+C(B-b)$$
As B=b=C-c, hence
$$b^2=b(C-c)$$
$$b=0,C-c$$

Comparing the constant terms,
$$bC+c=C(C-c)$$
$$bC+c=bC$$
$$c=0$$

From these, $b,c$ and $C$ are zero.

 

[ehild]

I made a sign error in my equation, I am afraid. It should be
b(Ax²+Bx+C)+c=(Ax²+Bx+C)(x²+bx+c-(Ax²+Bx+C)) or maybe not? You have to check what is said to you

ehild

 

[ehild]

Edited:

As it was my fault, I do one more step with the correct formula and delete anything further.

b(Ax²+Bx+C)+c=(Ax²+Bx+C)(x²(1-A)+(b-B)x+(c-C))

We know already that A=1 and B=b.
....

ehild

 

[Saitama]

As it was my fault, I repeat your derivation with the correct formula.

(Derivation)

Oh ehild, you should have waited for me to do it. I cannot reply instantly now a days because of my finals. I still haven't checked the derivation so you may wish to remove it. :)

 

[ehild]

I removed everything but a hint: before comparing further coefficients replace A=1 and B=b into the equation. Good luck for the finals! :)

ehild

 

[Saitama]

Good luck for the finals! :)

Thanks, my finals just got over today. I did well especially in Physics and Maths.

Getting back onto the question, I replaced A with 1 and B with b.
The equation reduces to
b(x<sub>2[/sup]+bx+C)+c=(x2[/sup]+bx+C)(c-C)

Comparing coefficients of x2[/sup],
b=c-C

Comparing coefficients of x,
b2[/sup]=(c-C)b or b2[/sup]=b2[/sup]. This means b=0?

Comparing the constant terms,
bC+c=C(c-C) or bC+c=Cb
Again c=0.

I am ending up with the same results again. :(</sub>

 

[ehild]

Thanks, my finals just got over today. I did well especially in Physics and Maths.

Congratulation! You are great!

Getting back onto the question, I replaced A with 1 and B with b.
The equation reduces to
b(x²+bx+C)+c=(x²+bx+C)(c-C)

Comparing coefficients of x²,
b=c-C

Comparing coefficients of x,
b²=(c-C)b or b²=b². This means b=0?

No, b²=b² does not mean anything for b. b is identical with itself as anything is identical with itself.

Comparing the constant terms,
bC+c=C(c-C) or bC+c=Cb
Again c=0.

Yes, c=0, so b=c-C means that C=-b.

I am ending up with the same results again. :(

It is not the same. Now you know that f has zero constant term, and you know the coefficients of g in terms of the coefficients of f :
A=1, B=b and C=-b.

Now use the equation g(2)=37.

ehild

 

[Saitama]

f(x)=x²+bx+c and g(x)=x²+bx-b

g(2)=37=4+2b-b
or b=33.

Hence, g(3)=9+3*33-33=75

Thanks a lot ehild!