Freaky Functions

$$f(g(x)) = f(x).g(x) \ \ (*)$$

Let $m=\deg f(x) ; \ n=\deg g(x)\ (m,n \in N)$ then from (*) we get $mn=m+n \Leftrightarrow m=n=0 \vee m=n=2$ .

If $m=n=0$ then $f(x), g(x)$ are constant, and since $g(2)=37$ , let $f(x)=\alpha$ we have $$\alpha = 37\alpha \Leftrightarrow \alpha=0$$ This is invalid since $f(x)$ is a non-zero polynomial. Therefore $m=n=2$ , which implies that $f(x)$ and $g(x)$ are quadratic polynomials.

Let $f(x)=ax^2+bx+c, g(x)=dx^2+ex+f \ \ (a,d \ne 0)$ . We expand both sides of (*):

$f(g(x))=a(dx^2+ex+f)^2+b(dx^2+ex+f)+c$

$f(x).g(x)=(ax^2+bx+c)(dx^2+ex+f)$

Comparing coefficients gives us

$af^2+bf+c=cf \ \ (1)$

$2aef + be=bf+ce \ \ (2)$

$ae^2+2adf+bd=af+cd+be \ \ (3)$

$2ade=ae+bd \ \ (4)$

$ad^2=ad \ \ (5)$

$(5) \Rightarrow d=1$ as $a,d\ne 0$ . $(4) \Rightarrow ae=b$ .

$(3)\Rightarrow be+2af+b=af+c+be \Leftrightarrow af=c-b$

$(1) \Rightarrow f(c-b)+bf+c=cf \Leftrightarrow c=0 \Rightarrow af=-b=-ae$

Therefore $e=-f$ . Additionally, $g(2)=4d+2e+f=4+2e+f=37$ , giving us $\fbox{d=1, e=33, f=-33}$ .

Thus $g(x)=x^2+33x-33 \Rightarrow g(3)=75$ .

Let f(x) be a polynomial of degree n and g(x) of degree m. Let y= f(g(x)-f(x).g(x)=0. The degree for f(g(x)= n m and that for f(x).g(x)=n+m. n+m=n m, then only the function y will be zero for all values of x.( f(g(x) and f(x).g(x) are identical functions, hence they are equal for all values of x). We get two solutions i.e n=m=0 and n=m=2. We cannot have n=m=0 because then the functions will become constant and g(x) will be equal to 1. But it is given g(2)=37,hence we cannot have n=m=0. Therefore, n=m=2. Let f(x)= ax^2+bx+c and g(x)= dx^2+ex+f. f(g(x)= a(dx^2+ex+f)^2+b(dx^2+ex+f)+c. f(x).g(x)= (ax^2+bx+c).(dx^2+ex+c).

Comparing coefficient of each term we get the following equations. 1. Equating coefficient of x^4 ,we get ad^2=ad. a cannot be zero since f(x) is a degree two polynomial and so cannot be d since g(x) is also degree two polynomial.Hence, d=1. 2. Equating coefficient of x^3, we get ae=b 3. Equating coefficient of x^2, we get af+b=c. 4. Equating the constant terms we get c=0. 5. By using g(2)=37, we get 2e+f=33.

By using equations 2,3,4,5 we get e+f=0. Hence, e=33 and f= -33. Therefore the polynomial g(x)= x^2+33x-33. g(3)=75.

Suppose $m$ and $n$ are the degree of $f(x)$ and $g(x)$ respectively, $m$ and $n$ are non-negative integers.

Since $f(g(x))=f(x).g(x)$ , $m.n=m+n$ . So $(m-1)(n-1)=1$ and we obtain $m=n=2$ .

Therefore, suppose $f(x)=(x-a)(x-b)$ and $g(x)=(x-c)(x-d)$ .

Since $f(g(x))=f(x).g(x)$ , we have $[(x-c)(x-d)-a][(x-c)(x-d)-b]$

$=(x-a)(x-b)(x-c)(x-d)$ .

Because $a$ and $b$ are equivalent, $c$ and $d$ are equivalent, we only need to consider 2 cases:

Case 1: $(x-c)(x-d)-a=(x-a)(x-c)$ and $(x-c)(x-d)-b=(x-b)(x-d)$ .

These will lead to $a=b=c=d=0$ and therefore not satisfy the given condition.

Case 2: $(x-c)(x-d)-a=(x-c)(x-d)$ and $(x-c)(x-d)-b=(x-a)(x-b)$ .

These will lead to $a=0$ and $cd=c+d=b$ .

Therefore, $g(x)=(x-c)(x-d)=x^2-bx+b$ . Since $g(2)=37$ , we obtain $b=-33$ and therefore $g(3)=75$ .

given: f(g(x)) = f(x) * g(x) .....(A) ; and g(2) = 37 .

to find: g(3)

Solution: Let degree of polynomial f(x)=m and that of g(x)=n where m&n are some whole number.

Since, Degree of f(g(x)) will be n * m and that of f(x) * g(x) = m + n ;

it implies that n * m = m + n ...(i) ; from this we have n = m * (n-1) which implies that n divides m, since n does not divide (n-1) for any natural number. Similarly m divides n.

Since m and n divide each other it means that m = n, which from equation (i) implies that either m = n = 2 or m = n = 0.

If m = n = 0 then f(x) and g(x) are both constant polynomials , say f(x) = p and g(x) = q for some real number p&q.

Then from equation (A) we have f(q) = p * q which implies p = p * q .

This means either p = 0 meaning f(x) is identically 0 which is not true (since f(x) is a non-zero polynomial) or q = 1 i.e. g(x) = 1 for all real values of x, which is also false since g(2) = 37.

Therefore m = n = 2 meaning f(x) and g(x) are quadratic polynomials.

Let f(x) = a * (x - x1) * (x - x2) ; and g(x) = u * (x - x3) * (x - x4) ;

i.e. x1 and x2 are roots of f(x) and x3 and x4 are roots of g(x).

Now, f(g(x)) = 0 if any of f(x) or g(x) is 0. If g(x) = 0 ; we have f(0) = f(x) * 0 = 0 i.e. 0 is a root of f(x). Let x2 = 0; so f(x) = a * x * (x - x1) ;

So, f(g(x)) = f(x) * g(x) = a * x *( x - x1) * u * (x - x3) * (x - x4) = a * u * x * (x - x1) * (x - x3) * (x - x4) ;

We have f(x) = a * x * (x - x1) ; Therefore , f(g(x)) = a * g(x) * (g(x) - x1) ...(ii) ; Since f(g(x)) = f(x) * g(x) ; from equation (A) and (ii) we get

a * g(x) * (g(x) - x1) = f(x) * g(x) which implies that

a * (g(x) - x1) = f(x) ; (since g(x) is not identically zero)

So, a * {u * (x-x3) * (x-x4) - x1} = a * x * (x - x1) ; which implies that

     u * (x - x3) * (x - x4) - x1 = x * (x - x1) ..... (iii) ;

because a is not equal to 0, since f(x) is a quadratic polynomial.

[Note: m^n denotes m to the power of n wherever it appears in the solution]

Expanding equation (iii) and shifting terms to one side we get

(u - 1) * (x^2) + {x1 - u * (x3 + x4)} * x + (u * x3 * x4 - x1) = 0 ... (iv) ;

Since, this equation is true for all real values of x, i.e. the equation is identically zero, the coefficients in the quadratic equation as well as the constant term must be zero.

Therefore u=1 ; x1 = x3 + x4 ; and x3 * x4 = x1 = x3 + x4 This implies that

g(x) = (x - x3) * (x - x4) ={ (x^2) - (x3 + x4) * x + x3 * x4 } = { (x^2) - x1 * x + x1 } .... (v) ;

Putting x= 2 in equation 5 we get, 37 = 4 - (2 * x1) + x1; implies x1 = (-33)

So, g(x) = (x^2) + 33 * x + (- 33) ... (vi)

Putting x = 3 in equation (vi) we get g(3) = 9 + 33 * 3 - 33 = 9 + 66 = 75

Hence, the solution is g(3) = 75 .

If $D(f)$ is the degree of $f$ etc. then we have: $D(f(g)) = D(f)D(g)$ and $D(fg) = D(f) + D(g)$ . These simultaneous equations only have two solutions in non-negative integers: $D(f) = D(g) = 2$ and $D(f) = D(g) = 0$ .

In the latter case, i.e. $f$ and $g$ both constant, the problem condition gives $f = fg$ , so $f=0$ or $g=1$ . But it is specified that $f \neq 0$ and $g = 37$ , so there is no solution here. Therefore $f,g$ are both quadratics.

If we multiply the two quadratics and equate like terms, it turns out (after some algebra that would take me too long to write out here) that there is only one general solution: $f(x) = ax(x+b) \\ g(x) = x(x+b) - b$ which is easily verifiable: $f(g) = ag(x(x+b)) = ag({f \over a}) = fg$ .

If $g(2) = 37$ then $b = 33$ , so $g(3) = \boxed{75}$ .

Let the degrees of polynomials $f$ and $g$ be positive integers $A$ and $B$ respectively, because $deg(LHS) = deg(RHS)$ , we have $AB = A + B$ , or $1 = \frac {1}{A} + \frac {1}{B}$ , so the solution is $A = B = 2$ . So we let $f(x) = ax^2 + bx + c, g(x) = dx^2 + ex + f$ , with $a \ne 0, d \ne 0$

Because $g(2) = 37$ , we get $2e + f = 33$ , $\space \space \space ((1))$

Given $f(g(x)) = f(x) \cdot g(x)$ , we have

$a(dx^2 + ex + f)^2 + b(dx^2 + ex + f) + c =$

$(ax^2 + bx + c)(dx^2 +ex + f)$

Expand and group them in terms of different powers of $x$

$x^4(ad^2) + x^3( 2ade) + x^2 (2adf+ae^2 +bd) + x(2aef$

$+be) + (af^2+ bf+c) =x^4 (ad) + x^3(ae+bd) +$

$x^2(af+be+cd) + x(bf+ce) + (cf)$

$\space \space \space \space \space \space ad^2 = ad \Rightarrow d = 1$ , $\space \space \space ((2))$

$\space \space \space \space \space \space ae = b$ , $\space \space \space ((3))$

$\space \space \space \space \space \space 2adf+ae^2 +bd = af+be+cd \Rightarrow af + b = c$ , $\space \space \space ((4))$

$\space \space \space \space \space \space af^2 + bf + c =cf \Rightarrow af^2 + bf+c = cf$ , $\space \space \space ((5))$

$((4)) \times f - ((5))$ : $\space \space \space \space c = 0 \Rightarrow af+ b = 0 \Rightarrow f = -\frac {b}{a}$ , $\space \space \space ((6))$

Letting $x=2$ for $f(g(x)) = f(x) \cdot g(x)$ , we get $\frac {f(37)}{f(2)} = 37$

$\space \space \space \space \space \space \large \Rightarrow \frac {1369a + 37b}{4a+2b} = 37$

$\space \space \space \space \space \space \large \Rightarrow \frac {1369 + 37b/a}{4+2b/a} = 37$

$\space \space \space \space \space \space \large \Rightarrow \frac {1369 -37f}{4-2f} = 37 \Rightarrow f = -33 \Rightarrow e = 33$

$\space \space \space \space \space \space \large \Rightarrow g(3) = 3^2 + 3 \cdot 33 - 33 = \boxed{75}$

Let the degree of $f(x)$ = $m$ and that of $g(x)$ = $n$ .Therefore the degree of $f(g(x))$ = $mn$ .Equating the degrees in the above relation that $f(g(x))$ = $f(x)$ . $g(x)$ we get $m$ = $\frac {n}{n-1}$ .Substituting $m$ in the equation $m+n=mn$ we get $n=2,m=2$ .Let us assume that $f(x)$ is a quadratic of the form $ax^2+bx+c=0$ .Putting $g(2)=37$ , we get $f(37)=37.f(2)$ .So in the quadratic form we get $a.37^2+b.37+c=37(a.2^2+b.2+c)$ .On simplification we get $1221a-37b-36c=0$ .The values of $a,b,c$ are $a=1,b=33,c=0$ .Therefore $f(x)=x^2+33x$ .Substituting this in $f(g(x))=f(x).g(x)$ we get $g(x)^2+33g(x)=(x^2+33).g(x)$ .This is a quadratic in $g(x)$ and from here we get the solutions $g(x)=0$ and $g(x)=x^2+33x-33$ .But $g(x)$ is a non-zero polynomial,therefore $g(x)=x^2+33x-33$ .Putting $x=3$ we get $g(3)=75$

Let f(x) = x^2 + cx + d and g(x) = x^2 + ax + b. Substituting this in the given equation yields (x^2 + ax + b)^2 + c(x^2 + ax + b) + d = (x^2 + ax + b)(x^2 + cx + d). By noting that the x^3, x^2, x, and constant coefficients must be equal, one finds that a = c, b + a = d, and d = 0. Putting everything in terms of a, one finds g(x) = x^2 + ax - a. Since g(2) = 37, it follows that a = 33. Then g(3) = 75.

let f(x) be a polynomial with degree m an g(x) with degree n. given:f(g(x))=f(x) g(x).............(1) f(g(x)) will be a polynomial with degree m^n.(Eg.if f(x)=x^2 and g(x)=x^3 then f(g(x))=x^6) f(x) g(x) will be a polynomial with degree m+n. then from (1),comparing coefficients of x on both sides m+n=m n i.e. m=n/(n-1) By theory of polynomials degrees m and n are positive integers By trial and error, m=n=2 is one solution therefore,the two polynomials are f(x)=x^2+b1 x+c1 g(x)=x^2+b2*x+c2

now substituting f(x) and g(x) in condition (1) and comparing coefficients (Whew!) we get, b1=b2=-c2 (hopefully.i don't have the courage to recheck :) ) therefore g(x)=x^2-c2 x+c2 but g(2)=37 therefore, c2=-33 finally,we get g(x)=x^2+33 x-33 now g(3)=3^3+33 3-33 * g(3)=75 **

First of all, let us assume f(x) be (a1)x^n + (b1)x^n-1 + c1x^n-2... and so on. similarly g(x) = (a2) x^m + b2. x^m-1 and so on... {where n and m are the degrees of f(x) and g(x) respectively.

Now substituting these values in f(g(x))=f(x)⋅g(x) and comparing the degrees of the new polynomial, we get:

mn = m+n. The only possible value of m and n is 2.

So that makes it f(x) = a.x^2 +bx+c and g(x) = px^2 + qx +c

As we know g(2) = 37. Therefore, using the relation given in the question, f(37) = f(2).g(2) = f(2).37

37^2.a + b.37 + c = 37( 4a +2b +c)

On simplifying we get, 37.33.a−37.b−36.c=0>

On inspection, possible values of a, b and c are 1, 33 and 0 respectively.

Therefore, f(x)=x^2+33.x

As f(g(x))=f(x).g(x)

(g(x))2+33⋅g(x)=(x^2+33.x)⋅g(x)

Solving the quadratic of g(x), we end up with g(x)=x^2 + 33.x − 33.

Now, substituting x=3, we get 3^2 + 33x3 - 33 = 9 + 99 - 33 = 75.

Considering the degrees of the polynomials on both sides of the equation we can determine that: $Deg(f(g(x))=Deg(f(x))*Deg(g(x))=Deg(f(x)*g(x))=Deg(f(x))+Deg(g(x))$

It follows that $Deg(f)*Deg(g)-Deg(f)-Deg(g)=0$ .

Adding $1$ to both sides we have that $Deg(f)*Deg(g)-Deg(f)-Deg(g)+1=1$ which implies that $(Deg(f)-1)(Deg(g)-1)=1$ .

As both $Deg(f)$ and $Deg(g)$ are integral we have the following two solutions: $Deg(f)=0, Deg(g)=0$ and $Deg(f)=2, Deg(g)=2$ .

We can discard the first because then $f$ and $g$ would be constant and we would have that $a=ab$ where $f(x)=a$ and $g(x)=b$ , but by hypothesis $b=37$ . So we would have $a=37a$ which implies $a=0$ , but we assumed that $f$ was non-zero so this case leads to contradiction.

Instead both $f$ and $g$ must be quadratic. It is now very tempting to try to write out the equation we were given in terms of coefficients and match up both sides but if we think about the properties of the given equation a bit more before we do that it will ease the difficulty of the computation tremendously.

First, note that if $g$ has a root $r$ then the given equation implies that $f(g(r))=f(r)*g(r)$ which implies that $f(0)=f(r)*0=0$ so the constant term of $f$ is $0$ . Also wlog we can assume f is monic because if not we can divide both sides by its leading term preserving equality (and because this procedure does not affect $g(x)$ )

Now we plug in $x=2$ , which yields that $f(g(2))=f(37)=f(2)*g(2)=f(2)*37)$ . If we let $f(x)=x^2+bx$ we have that $37^2+37b = 148+74b$ , so solving we have $b=33$ .

Finally we plug in $x=3$ , this yields that $f(g(3))=f(3)*g(3)$ . Treating $g(3)$ as a variable we can plug into the equation for $f$ to get $g(3)^2+33g(3)=(3^2+33*3)g(3)=108g(3)$ .

Collecting terms, we see that $g(3)^2-75g(3)=0$ . This implies that $g(3)(g(3)-75)=0$

Therefore either $g(3)=0$ or $g(3)=75$ .

If we let the leading coefficient of $g$ be $c$ then from the given equation and our knowledge of $f$ we see that $c^2=c$ and as $Deg(g)=2$ , $c$ cannot be $0$ , so $c=1$ . But then $g(x)$ can be factored as $(x-d)(x-e)$ .

Suppose $g(3)=0$ , then $g(x)=(x-3)(x-e)$ , plugging in $x=2$ we get that $37=-1*(2-e)$ so $e=39$ , but we can check that in this case the given equation does not hold.

It follows that $g(3)=\boxed{75}$

Let be $f(x)$ a polynomial of degree $n$ and $g(x)$ a polynomial of degree $m$ . Then $f(x) \cdot g(x)$ has degree $m + n$ and $f(g(x))$ has degree $m \cdot n$ . Since $f(g(x)) = f(x) \cdot g(x)$ we have:

$m \cdot n = m + n$ .

There are 2 possible solutions:

1. $n = m = 0$ .
2. $m = n = 2$ .

In case 1. we have $f(x)$ and $g(x)$ constant polynomials. Let be $f(x) = k$ and $g(x) = c$ , then we have: $k = c \cdot k \iff c = 1, \forall k \in \mathbb{R}$ But since we know that $g(2) = 37$ $g(x) = 1$ can't be the solution.

Case 2. To simplify calculations we can also notice that $0$ is a root of the polynomial $f(x)$ . This can be proved observing that if $x_0$ is a root of $g(x)$ then we have:

$f(g(x_0)) = f(x_0) \cdot g(x_0) \iff f(0) = 0$

Then we have, without losing generality:

$f(x) = kx(x - f_0); g(x) = c(x - g_0) ( x - g_1)$

$f(g(x)) = kc(x - g_0) (x - g_1) (c (x - g_0) (x - g_1) - f_0) =$

$f(x) \cdot g(x) = kcx (x - f_0) (x - g_0) (x - g_1)$

For $x = g_0$ or $x = g_1$ the identity $f(g(x)) = f(x) \cdot g(x)$ is true, therefore we can divide by $(x - g_0) (x - g_1)$ obtaining:

$x^2 - f_0x = cx^2 - c(g_1 + g_0)x + c g_1 g_0 - f_0$

then we have:

$c = 1$

$g_1 + g_0 = -f_0$

$g_1 g_0 = f_0$

( $g_0, g_1 \in \ \mathbb{C}, f_0 \in \mathbb{R}$ ) Then

$g(x) = x^2 +f_0x - f_0$

$g(2) = 37 \iff f_0 = 33$

$g(3) = 75$