Find the domain of $g\circ f$ where $f(x)=\frac{x}{x-64}$ and $g(x)=\sqrt{x}$ .

A) $(0,64]\cup (64,\infty )$
B) $(-\infty ,0]\cup (64,\infty )$
C) $[0,8)\cup (8,\infty )$
D) $(0,\infty )$
E) $(0,\infty )$ .

The domain of $g\circ f$ is determined by the restrictions imposed by both $f(x)$ and $g(x)$ . For $f(x)=\frac{x}{x-64}$ , the only restriction is that $x\ne 64$ to avoid division by zero. For $g(x)=\sqrt{x}$ , $x$ must be non-negative since we cannot take the square root of a negative number in the real number system. When we compose $g$ with $f$ , we need the output of $f(x)$ to be non-negative, which happens for $x>64$ and $x<0$ . However, since $x$ cannot be 64, the domain of $g\circ f$ is $(-\infty ,0]\cup (64,\infty )$ .