Be functional

Let $f(-1) = a$ and $f(0) = b$ .

Case 1: $a \ne b$ . Then $f(x) = (x-a)(x-b)$ , and we have $(a+1)(b+1)=a$ and $ab=b$ . So $a = 1$ or $b = 0$ . If $b = 0$ we get $a+1 = a$ , impossible. If $a = 1$ we get $2(b+1)=1$ , so $b = -1/2$ . So $f(x) = (x-1)(x+1/2)$ .

Case 2: $a = b$ . Then $f(x) = (x-a)(x-c)$ for some $c$ , and we have $(a+1)(c+1) = a$ and $ac = a$ . So $c = 1$ or $a = 0$ . If $a = 0$ we get $c+1 = 0$ , so $c = -1$ and $f(x) = x(x+1)$ . If $c = 1$ we get $2(a+1) = a$ , so $a = -2$ and $f(x) = (x+2)(x-1)$ .

So there are a total of three polynomials in the collection, and the sum of $f(3)$ for each of these is $7+12+10 = \fbox{29}$ .

Let $f(x) = x^2 + ax + b$ , then $f(0) = b, f(-1) = 1- a + b$ .

$f(f(0)) = 0 \iff f(b) = 0 \iff b^2 + ab + b = 0$

$\implies b = 0 \text{ or } b + a + 1 = 0$

Let statement $p$ be $“b = 0”$ and statement $q$ be $“b + a + 1 = 0”$ .

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$\phantom{\iff} f(f(-1)) = 0$

$\iff f(1 - a + b) = 0$

$\iff (1- a + b) ^2 + (1 - a + b)a + b = 0$

$\iff 1 + a^2 + \color{#3D99F6}{b^2} -2a +2b - 2ab + a -a^2 + \color{#3D99F6}{ab + b} =0 ~~~ \color{#3D99F6}{[b^2 + ab + b = 0]}$

$\iff 1- a + 2b - 2ab = 0$

$\iff (1-a)(2b + 1) = 0$

$\implies a = 1 \text{ or } b = -\frac{1}{2}$

Let statement $r$ be $“a = 1”$ and statement $s$ be $“b = -\frac{1}{2}”$ .

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We want $(p \vee q) \wedge (r \vee s)$ to be true. This is possible when at least one of $p$ and $q$ and at least one of $r$ and $s$ are true. All the possible cases are enumerated below. A dash $(-)$ represents a contradiction.

$\begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & s && a & b\\ \hline \text{T} & \text{F} & \text{T} & \text{F} && 1 & 0\\ \hline \text{T} & \text{F} & \text{F} & \text{T} && - & -\\ \hline \text{T} & \text{F} & \text{T} & \text{T} && - & -\\ \hline\\ \hline \text{F} & \text{T} & \text{T} & \text{F} && 1 & -2\\ \hline \text{F} & \text{T} & \text{F} & \text{T} && -\frac{1}{2} & -\frac{1}{2}\\ \hline \text{F} & \text{T} & \text{T} & \text{T} && - & -\\ \hline\\ \hline \text{T} & \text{T} & \text{T} & \text{F} && - & -\\ \hline \text{T} & \text{T} & \text{F} & \text{T} && - & -\\ \hline \text{T} & \text{T} & \text{T} & \text{T} && - & -\\ \hline\end{array}$

Three ordered pairs of $(a, b)$ are obtained viz. $(1, 0), (1, -2), (\frac{1}{2}, -\frac{1}{2})$ . Therefore the three quadratic equations are $x^2 + x, x^2 + x - 2, x^2 -\frac{1}{2}x - \frac{1}{2}$ .

Substituting $x =3$ gives $f(3) = 12, 10, 7$ respectively.

The sum of all possible values of $f(3)$ is $12 + 10 + 7 = \boxed{29}$ . $_\square$

Define the quadratic symbolically. Form two expressions representing $f(f(-1))$ and $f(f(0))$ and pass them to a solver, which will pass pack pairs of solutions for $b$ and $c$ , one at a time. For each solution pair, substitute the parameters in the quadratic as well as the value of the indeterminate, $x$ to obtain the value $f(3)$ . Keep running total.

We can split this problem into two cases, where $f(0)=0$ and where $f(0) \neq 0$ .

Case 1: $f(0)=0$ . We have $f(x)=x^2+bx$ . Now $f(-1)=1-b$ and $f(f(-1))=(1-b)^2+b(1-b)=1-b$ . Since $f(f(-1))=0$ , $1-b=0$ and $b=1$ giving $f(x)=x^2+x$ and $f(3)=12$ .

Case 2: $f(0) \neq 0$ . Let $f(0)=c$ so $f(c)=0$ . Let $f(x)=x^2+bx+c$ . From Vieta's, the product of the roots is $c$ , so the other root must be $1$ , and $b=-(c+1)$ . We now have $f(x)=x^2-(c+1)x+c$ . Plugging in $-1$ , we get $f(-1)=2+2c$ . Since the only roots are $c$ and $1$ , we either have $f(-1)=c$ or $f(-1)=1$ . If $f(-1)=c$ , we get $2+2c=c$ and $c=-2$ , giving $f(x)=x^2+x-2$ and $f(3)=10$ . If $f(-1)=1$ , we get $2+2c=1$ and $c=-1/2$ , giving $f(x)=x^2-\frac{x}{2}-\frac{1}{2}$ and $f(3)=7$ .

The values of $f(3)$ are $7,10,12$ and the sum is $\boxed{29}$ .