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) $( 0 , \infty )$.

Question

Quizplus · Accepted Answer

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 $x \neq 64$ 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 $(-\infty, 0] \cup (64, \infty)$.

Find the Domain Of g∘fg \circ fg∘f Where f(x)=xx−64f ( x ) = \frac { x } { x - 64 }f(x)=x−64x​

Find the Domain Of $g \circ f$ Where $f ( x ) = \frac { x } { x - 64 }$