Composing functions

Denote the first equation by $(1)$ . Let $max(a, b)$ denote the maximum value of a and b. Let the degrees of $f(x)$ and $g(x)$ be $f$ and $g$ respectively. Note that $f > 0$ since $f(0) \ne f(1)$ . By $(1)$ , $fg = max(f, g)$ . We now have two cases.

i) $g \ge f$

Then $fg = g$ . Now we have two cases.

a) $g = 0$

Then $g(x) = c$ for some $c \in \mathbb{R}$ . But, by $(1)$ , $f(x) = f(c) - c$ , which is a constant, which contradicts $f > 0$ .

b) $f = 1$

Then $f(0) = 5$ and $f(1) = 7$ implies $f(x) = 2x + 5$ . Substituting this into $(1)$ gives $g(x) = 2x$ , which contradicts $g(0) \ne 0$ .

ii) $f > g$ Then, $fg = f$ . Since $f > 0$ , $g = 1$ . Let $g(x) = ax + b$ . Let $f(x) = a_{n}x^{n} + a_{n - 1}x^{n - 1} + ... + a_{0}$ , $a_{n} \ne 0$ . The leading coefficient of $f(g(x))$ is $a^{n} \cdot a_{n}$ and the leading coefficient of $f(x) + g(x)$ is $a_{n}$ .  Thus, $a^{n} = 1$ . We again have two cases.

a) $a = 1$

Then, since $g(0) \ne 0$ , $b \ne 0$ . Equation $(1)$ gives $f(x + b) - f(x) = x + b$ . The degree of the LHS is $n - 1$ while the degree of the RHS is $1$ , so $n = 2$ . Note that $a_{0} = f(0) = 5$ and $a_{1} + a_{2} = f(1) - a_{0} = 2$ . Substituting this into $f(x + b) - f(x) = x + b$ and solving for $a_{1}, a_{2}$ and $b$ gives $a_{1} = \frac{1}{2}, a_{2} = \frac{3}{2}, b = \frac{1}{3}$ . It remains to note that the set of polynomials $f(x) = \frac{3}{2}x^{2} + \frac{1}{2}x + 5, g(x) = x + \frac{1}{3}$ satisfies the equations. This gives $f(20) = 615$ .

b) $a = -1$ and $n$ is even.

Again, $f(-x + b) - f(x) = -x + b$ and the degree of LHS is still $n - 1$ since $n$ is even. Thus, $n = 2$ using the same argument as the above. Using similar arguments, we can find the equations $a_{1} + a_{2} = 2$ , $-2ba_{2} - 2a_{1} = -1$ , $b^{2}a_{2} + ba_{1} = b$ . Simplifying the last two equations give $ba_{2} + a_{1} = \frac{1}{2}$ and $ba_{2} + a_{1} = 1$ , which is a contradiction. Thus, $f(20) = 615$ is the only solution.

Let $d_f$ and $d_g$ be the degrees of $f(x)$ and $g(x)$ , respectively, and let $f$ and $g$ be the leading coefficients of $f(x)$ and $g(x)$ , respectively.

Note that $f(x)$ is not a constant, since $f(0)$ and $f(1)$ take on different values. If $g(x)$ is constant, then let $g(x)=c$ . Then, we have

$\begin{aligned} f(c)&=f(x)+c\\ \implies f(x)&=f(c)-c, \end{aligned}$

but $f(c)-c$ is constant since $c$ is a fixed value, so hence we get $f(x)$ is constant, a contradiction. Hence, we have that $d_f,d_g\ge 1$ .

Note that the degree of $f(g(x))$ is $d_fd_g$ , and the degree of $f(x)+g(x)$ is at most $\max(d_f,d_g)$ (The reason why we have the "max" is because if $d_f=d_g$ and $f=-g$ , then we don't have equality). From $f(g(x))=f(x)+g(x)$ , we get that $d_fd_g\le \max(d_f,d_g)$ .

If $d_f\ge d_g$ , then we get that

$\begin{aligned} d_fd_g&\le d_f\\ \implies d_g&\le 1, \end{aligned}$

so $d_g=1$ since $g(x)$ is not constant.

Similarly, if $d_g\ge d_f$ , we get that $d_f=1$ .

Hence, either $f(x)$ or $g(x)$ is linear.

If $f(x)$ is linear, then from $f(0)=5$ and $f(1)=7$ we get that $f(x)=2x+5$ . Plugging this into $f(g(x))=f(x)+g(x)$ , we get that

$\begin{aligned} 2g(x)+5&=g(x)+2x+5\\ \implies g(x)&=2x, \end{aligned}$

but this means that $g(0)=0$ , contradicting $g(0)\neq 0$ .

Thus, we have that $g(x)$ is linear, and $f(x)$ has degree at least $2$ . Let $g(x)=ax+b$ .

Then, we get that $f(ax+b)=f(x)+ax+b$ . Now, we compare the leading coefficients of both sides of this equation.

The leading coefficient of the left side is $a^{d_f}f$ , and the leading coefficient of the right side is $f$ since $f(x)$ has degree at least $2$ , so $ax+b$ doesn't affect the leading coefficient. Hence,

$\begin{aligned} a^{d_f}f&=f\\ \implies a^{d_f}&=1\\ \implies a&=\pm 1. \end{aligned}$

If $a=-1$ , then we get that $f(-x+b)=f(x)-x+b$ . Plugging in $x=b$ gives $f(0)=f(b)=5$ , so define $p(x)=f(x)-5$ . Then, we get that $p(0)=p(b)=0$ . Note that since $g(0)\neq 0$ , we get that $b\neq 0$ . Since the degree of $p(x)$ is at least $2$ , we have that $p(x)=x(x-b)q(x)$ for some polynomial $q(x)$ .

We have that

$\begin{aligned} f(-x+b)&=f(x)-x+b\\ \implies f(-x+b)-5&=(f(x)-5)-x+b\\ \implies p(-x+b)&=p(x)-x+b\\ \implies (-x+b)(-x)q(-x+b)&=x(x-b)q(x)+(-x+b)\\ \implies (-x)q(-x+b)&=-xq(x)+1\\ \implies x(q(x)-q(-x+b))&=1 \end{aligned}$

If $q(x)=q(-x+b)$ , then we get that $0=1$ , a contradiction. Otherwise, we have that the left side has degree at least $1$ while the right side is constant, a contradiction.

Hence, we can't have $g(x)=-x+b$ , so we must have $g(x)=x+b$ . Unfortunately, a similar solution doesn't work for this case.

We have that $f(x+b)=f(x)+x+b$ , or $f(x+b)-f(x)=x+b$ . But $f(x+b)-f(x)$ is just a finite difference of $f(x)$ , so hence it has degree $d_f-1$ (Also, after expanding $f(x+b)-f(x)$ , it is easy to see that the coefficient of $x^{d_f}$ is $0$ and the coefficient of $x^{d_f-1}$ is $bfd_f$ , which is nonzero since $b,f,d_f\neq 0$ ).

Hence, the left side of $f(x+b)-f(x)=x+b$ has degree $d_f-1$ and the right side has degree $1$ , so hence we have $d_f=2$ and thus $f(x)$ is quadratic.

Let $f(x)=rx^2+sx+t$ . Since $f(0)=5$ we have $f(x)=rx^2+sx+5$ . Since $f(1)=7$ we have that $r+s=2$ and $f(x)=rx^2+(2-r)x+5$ . Plugging this into $f(x+b)-f(x)=x+b$ gives

$\begin{aligned} [r(x+b)^2+(2-r)(x+b)+5]-[rx^2+(2-r)x+5]&=x+b\\ \implies 2rbx+rb^2+(2-r)b&=x+b, \end{aligned}$

so equating coefficients gives us $2rb=1\implies rb=\frac 12$ and

$\begin{aligned} rb^2+(2-r)b&=b\\ \implies rb+2-r&=1\\ \implies \frac 12+2-r&=1\\ \implies r&=\frac 32, \end{aligned}$

where we divided by $b\neq 0$ and substituted $rb=\frac 12$ . Then, we get that $f(x)=\frac 32 x^2+\frac 12 x+5$ and $g(x)=x+\frac 13$ . It is easy to check that these two polynomials satisfy our original conditions, so our only possibility for $f(20)$ is $\frac 32 \cdot 20^2+\frac 12 \cdot 20+5=\boxed{615}$ .

Consider the following, where $\deg(f(x))$ denotes the degree of $f(x)$ . $\deg(f(g(x)))=\deg(f(x)+g(x))$ $\deg(f(x))\cdot\deg(g(x))=\max(\deg(f(x)),\deg(g(x)))$ Therefore, one of $(f(x),g(x))$ is linear. First, assume $f(x)$ is linear. Therefore, $f(x)=2x+5$ Plugging into the first equation: $2g(x)+5=2x+5+g(x)$ $g(x)=2x$ However, since $2(0)=0$ , this does not satisfy the fourth given condition. Therefore, $\deg(f(x))\ge 2$ and $\deg(g(x))=1$ . Let $g(x)=ax+b$ , and $b\neq 0$ by the fourth given condition. $f(ax+b)=f(x)+ax+b$ Considering the leading term of both sides, $a=1$ . $f(x+b)-f(x)=x+b$ This expression is an algebraic identity. Let $f(x)=a_1x^n+a_2x^{n-1}\cdots$ Now, the coefficient of the term with degree $n-1$ is $a_1nb$ on the LHS and either 0 or 1 on the right side. Therefore, $n=2$ , since $a_1,n,b\neq 0$ . Now, let $f(x)=a_1x^2+a_2x+5$ . We can expand and equate coefficients, giving: $a_1b=\dfrac{1}{2}$ $a_1b+a_2=1$ Therefore, $a_2=\dfrac{1}{2}$ . $f(x)=a_1x^2+\dfrac{1}{2}x+5$ Since $f(1)=7$ , we find that $a_1=\dfrac{3}{2}$ . We now know only one polynomial $f(x)$ satisfies the conditions. The answer is $f(20)=\boxed{615}$ .

First, let the degree of f be i and the degree of g be j. We see that i*j=max(i,j).

Suppose j>=i. We see that i*j=j, so, assuming g is nonconstant, i=1. This implies f=2x+5. Therefore, 2g+5=2x+5+g, so g=2x, which is impossible. If j=0, then we have that i=0, contradiction, since f is nonconstant.

Therefore, i>j. Since the first condition gives us that i*j=i. Since i>0 (because f is nonconstant), we have that j=1. Now, consider the leading coefficient of f(g(x)). We know it must equal the leading coefficient of f(x), so we see that g(x)=x+y, for some nonzero constant y. Furthermore, (x+y)^i has positive coefficients of degree i-1, so it follows that i=2.

Therefore f(x)=ax^2+bx+5. Substituting into our equation, ax^2+2ayx+ay^2+bx+by+5=ax^2+bx+5+x+y, so 2ayx+by+ay^2=x+y. Matching coefficients, 2ay=1, so ay=1/2. Furthermore, ay^2+by=y. Since y is nonzero, ay+b=1 and b=1/2.

Now we have f(x)=ax^2+x/2+5. We see that a=3/2 by the third initial condition, so the answer is

3/2 400+1/2 20+5=615