What's the opposite of a zero?

Let:

$f(x)=\left\lfloor\dfrac{1}{x^2-6x+10}\right\rfloor$

$g(x)=\left\lfloor\dfrac{x^2+x}{x^4+3x^2+x+1}\right\rfloor$

$h(x)=\left\lfloor\dfrac{1}{x^2+x+1}\right\rfloor$

$$ $$

We can rewrite $\dfrac{1}{x^2-6x+10}$ as $\dfrac{1}{(x-3)^2+1}$

We can see that for all $x\not=3$ , $\dfrac{1}{x^2-6x+10}>0$ as $(x-3)^2+1 > 1$

At $x=3$ , $\dfrac{1}{x^2-6x+10} = 1$ as $(x-3)^2+1=1$

Therefore, $f(x) = \left\{\begin{array}{lr} 0 & x\not=3 \\ 1 & x=3 \end{array}\right\}$

$$ $$

We can rewrite $\dfrac{x^2+x}{x^4+3x^2+x+1}$ as $\dfrac{x(x+1)}{(x^2+1)^2+x(x+1)}$

We can see that the zeros occur at $x=-1$ and $x=0$ , with the function negative over $-1<x<0$

Setting equal to 1:

$\dfrac{x(x+1)}{(x^2+1)^2+x(x+1)} = 1$

$x(x+1) = (x^2+1)^2+x(x+1)$

$(x^2+1)^2 = 0$

No solution

$$

Setting equal to -1:

$\dfrac{x(x+1)}{(x^2+1)^2+x(x+1)} = -1$

$x(x+1) = -(x^2+1)^2-x(x+1)$

$(x^2+1)^2 +2x(x+1) = 0$

Minimum of $(x^2+1)^2$ is 1, and minimum of $2x(x+1)$ is $-\frac{1}{2}$ , therefore:

No solution

$$

We now know that $-1<\dfrac{x^2+x}{x^4+3x^2+x+1}<1$

Therefore, $g(x) = \left\{ \begin{array}{lr} 0 & x\le-1 \\ -1 & -1<x<0 \\ 0 & x\ge0 \end{array}\right\}$

$$ $$

Finally,

$\dfrac{1}{x^2+x+1} > 0$ for all x

$$

Setting equal to 1:

$\dfrac{1}{x^2+x+1} = 1$

$x^2+x+1 = 1$

$x^2 + x = 0$

$x=-1$ and $x=0$

$$

Setting equal to 2:

$\dfrac{1}{x^2+x+1} = 2$

$2x^2+2x+2 = 1$

$2x^2+2x+1 = 0$

Minimum of $2x^2+2x+1$ is $\frac{1}{2}$ , therefore,

No solution

$$

Therefore, $h(x) = \left\{\begin{array}{lr} 0 & x<-1 \\ 1 & -1\le x \le 0 \\ 0 & x>0 \end{array}\right\}$

$$ $$

Looking at these three functions we can see that their sum is equal to 0 everywhere except $x=-1$ , $x=0$ , and $x=3$ , where it's equal to 1.

Therefore, the answer is $-1 + 0 + 3 = \fbox{2}$