Find the domain of the function $$f(x) = \frac{1}{(x-2)^2+1/(x-8)}.$$ Express your answer in interval notation.
To find the domain of the function, we need to identify any values of $x$ that would make the function undefined. In this case, the function will be undefined if the denominator is equal to zero, since division by zero is undefined.
The denominator is $(x-2)^2 + \frac{1}{x-8}$. For this expression to be zero, both terms in the denominator would have to be zero simultaneously: $(x-2)^2 = 0$ and $\frac{1}{x-8} = 0$.
Solving the first equation, $(x-2)^2 = 0$, we find that $x-2 = 0$, which gives $x = 2$.
Solving the second equation, $\frac{1}{x-8} = 0$, we find that the denominator can never be zero since the numerator is a constant value of 1.
Therefore, the only value of $x$ that makes the denominator zero is $x = 2$.
Hence, the domain of the function is all real numbers $x$ except $x = 2$. In interval notation, we can write this as $(-\infty, 2) \cup (2, \infty)$.
We start by examining the denominator of this function, which is $(x-2)^2+1/(x-8)$. Let us consider when this expression is equal to 0.
The expression $(x-2)^2$ is never equal to zero, so we only need to consider when $1/(x-8) = 0$. This equation holds when the numerator is equal to zero: $$x-8 = 0.$$Thus, the denominator is equal to zero when $x=8$.
However, since the denominator does not appear directly in the original function, we need to consider whether the denominator could be zero when the function is defined. Since $1/(x-8)$ only "checks" whether $x-8$ is zero, we only need to check if $(x-2)^2$ could be equal to 0.
The expression $(x-2)^2$ is equal to zero exactly when $x-2 = 0$, which means $x=2$. Therefore, the only value of $x$ for which the denominator is equal to zero is $x=8$, and the function is not defined when $x=8$.
Thus, the domain of the function $f(x) = \frac{1}{(x-2)^2+1/(x-8)}$ is the set of all real numbers except $x=8$, or $$\boxed{(-\infty, 8) \cup (8,\infty)}$$ in interval notation.
To find the domain of a function, we need to determine the set of all possible values that x can take. In this case, we need to consider any restrictions that would make the function undefined.
In this function, we have two potential restrictions. The first one occurs when the denominator of the fraction becomes zero. So, let's find the values of x when $(x-2)^2 + \frac{1}{x-8} = 0$.
To solve this equation, we can start by simplifying the equation as much as possible.
$(x-2)^2 + \frac{1}{x-8} = 0$
Expanding the square, we get:
$x^2 - 4x + 4 + \frac{1}{x-8} = 0$
Now, let's multiply through by $(x-8)$ to eliminate the fraction:
$(x^2 - 4x + 4)(x-8) + 1 = 0$
Expanding and simplifying the equation further, we get:
$x^3 - 12x^2 + 27x - 135 = 0$
Now, we need to solve this cubic equation. However, finding the exact values of x might not be feasible without the assistance of numerical methods or calculators.
So, let's rely on the rational root theorem to identify potential rational roots. The rational root theorem states that any rational root of the polynomial equation with integer coefficients can be written in the form of p/q, where p is a factor of the constant term (in this case, 135), and q is a factor of the leading coefficient (in this case, 1).
By trying out various combinations of p and q that satisfy these conditions, we can find a list of potential rational roots. Then, we can test each of these values using synthetic division or a similar method to see if they are roots of the equation.
Once we have found all the real roots of the equation (if any), we can identify whether any of them fall within the domain of the function.
Unfortunately, this process can be time-consuming and complex. However, we can still determine the domain of the function by considering the overall behavior of the function.
Notice that the expression $(x-2)^2$ is always non-negative since it involves squaring a real number. Additionally, the expression $\frac{1}{x-8}$ becomes undefined when $x = 8$. Therefore, the function $f(x)$ will only be undefined when the denominator $(x-2)^2 + \frac{1}{x-8}$ becomes zero.
Hence, the domain of $f(x)$ is the set of all real numbers except $x = 8$, as there are no other restrictions on x.
Therefore, the domain of the function $f(x)$ is $(-\infty, 8) \cup (8, \infty)$.