Find the domain of the function $$f(x) = \frac{1}{(x-2)^2+1/(x-8)}.$$ Express your answer in interval notation.

Answer 1

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.