Let $f ( x ) = \sqrt { x - 1 }$ and $g ( x ) = \frac { 2 } { x }$ Then the domain of $g \circ f$ is
A) $[ 1 , \infty )$
B) $( 1 , \infty )$
C) $( 0 , \infty )$
D) $( 0,1 ) \cup ( 1 , \infty )$
E) $[ 0 , \infty )$

Question

Quizplus · Accepted Answer

The domain of $g \circ f$ is determined by first considering the domain of $f(x) = \sqrt{x - 1}$, which is $[1, \infty)$ because the expression under the square root must be non-negative. Then, for $g(f(x)) = \frac{2}{\sqrt{x - 1}}$, the denominator cannot be zero, which excludes $x = 1$ from the domain of $g \circ f$. Therefore, the domain of $g \circ f$ is $(1, \infty)$.

Let f(x)=x−1f ( x ) = \sqrt { x - 1 }f(x)=x−1​

Let $f ( x ) = \sqrt { x - 1 }$