Recall that √x has a domain of x ≥ 0.
So, f(x) is defined as long as
(x + 1)/(x - 1) ≥ 0
• We have equality when x = -1
• Otherwise (x + 1)/(x - 1) is positive if both x + 1 and x - 1 are positive, or both are negative:
[tex]\begin{cases}x+1>0 \implies x>-1 \\ x-1>0 \implies x>1\end{cases} \implies x > 1[/tex]
[tex]\begin{cases}x+1<0 \implies x<-1 \\ x-1<0 \implies x<1\end{cases} \implies x<-1[/tex]
Then the domain of f(x) is
x > 1 or x ≤ -1
On the other hand, g(x) is defined by two individual square root expressions with respective domains of
• x + 1 ≥ 0 ⇒ x ≥ -1
• x - 1 ≥ 0 ⇒ x ≥ 1
but note that g(1) is undefined, so we omit it from the second domain.
Then g(x) is defined so long as both x ≥ -1 *and* x > 1 are satisfied, which means its domain is
x > 1
f(x) and g(x) have different domains, so they are not the same function.
