Find the sum of all real numbers $x$ that are not in the domain of the function $$f(x) = \frac{1}{x^2+7} + \frac{1}{x^3 - x^4} + \frac{1}{x^2 - 3x + 2}.$$
The function is defined unless a denominator is equal to zero. The first denominator, $x^2+7$, is never zero. The second denominator, $x^3 - x^4 = x^3(1 - x)$, is zero either when $x = 0$ or $x = 1$. The third denominator is zero when $x = 2$ or $x = 1$. Thus the function is defined for all $x$ except $x = 0, 1, 2$. The sum of these three real numbers is $0+1+2=\boxed{3}.$
To find the values of $x$ that are not in the domain of the given function, we need to look for any values of $x$ that would make the denominator equal to zero.
We check the first denominator $x^2+7$. Setting $x^2+7=0$, we find that there are no real values of $x$ that satisfy this equation. So, the first fraction $\frac{1}{x^2+7}$ is defined for all real numbers.
Next, we consider the second denominator $x^3-x^4$. Setting $x^3-x^4=0$, we can factor out an $x$: $x(x^2-x^3)=0$. This equation has two solutions: $x=0$ and $x^2-x^3=0$. The quadratic equation $x^2-x^3=0$ can be rewritten as $x(x-1)=0$, which gives two additional solutions: $x=0$ and $x=1$. Therefore, the second fraction $\frac{1}{x^3-x^4}$ is not defined at $x=0$ and $x=1$.
Lastly, we examine the third denominator $x^2-3x+2$. Setting $x^2-3x+2=0$, we can factorize it as $(x-1)(x-2)=0$. This equation has two solutions: $x=1$ and $x=2$. Hence, the third fraction $\frac{1}{x^2-3x+2}$ is not defined at $x=1$ and $x=2$.
In conclusion, the values of $x$ that are not in the domain of the function are $x=0$, $x=1$, and $x=2$.
To find the sum of all these real numbers, we simply add them up:
$$0 + 1 + 2 = 3.$$
Therefore, the sum of all real numbers $x$ that are not in the domain of the given function is $3$.
To find the sum of all real numbers $x$ that are not in the domain of the function $f(x)$, we need to determine where $f(x)$ is undefined.
First, let's analyze each term in $f(x)$ separately to identify any potential restrictions on the domain:
1. The term $\frac{1}{x^2+7}$ is defined for all real numbers $x$ since the denominator $x^2+7$ is always positive ($x^2$ is always non-negative, and adding 7 makes it positive).
2. The term $\frac{1}{x^3 - x^4}$ is defined for all real numbers $x$ except when the denominator $x^3 - x^4$ is equal to zero. To find the values of $x$ that make the denominator zero, we set $x^3 - x^4 = 0$ and solve for $x$.
If we factor out $x^3$ from the equation, we have $x^3(1 - x) = 0$. Using the zero-product property, we get $x^3 = 0$ or $1 - x = 0$. Solving these equations gives us $x = 0$ or $x = 1$. So, the function is undefined for $x = 0$ and $x = 1$.
3. The term $\frac{1}{x^2 - 3x + 2}$ is defined for all real numbers $x$ except when the denominator $x^2 - 3x + 2$ is equal to zero. To find the values of $x$ that make the denominator zero, we set $x^2 - 3x + 2 = 0$ and solve for $x$.
We can factor the quadratic equation as $(x - 1)(x - 2) = 0$. Using the zero-product property, we obtain $x - 1 = 0$ or $x - 2 = 0$, which gives $x = 1$ or $x = 2$. So, the function is undefined for $x = 1$ and $x = 2$.
Combining all the restrictions, we find that the function $f(x)$ is undefined for $x = 0, 1,$ and $2$. Therefore, the sum of all real numbers $x$ that are not in the domain of $f(x)$ is $0 + 1 + 2 = \boxed{3}$.