Find the Domain of the Function
The domain of the function
\[ f(x)=\sqrt{4-x}+\frac1{\sqrt{x^2-1}} \]
is equal to
(a) \((-\infty,-1)\cup(1,4)\)
(b) \((-\infty,-1]\cup(1,4]\)
(c) \((-\infty,-1)\cup[1,4]\)
(d) \((-\infty,-1)\cup[1,4)\)
For
\[ \sqrt{4-x} \]
we need
\[ 4-x\ge0 \]
\[ x\le4 \]
For
\[ \frac1{\sqrt{x^2-1}} \]
denominator must be positive:
\[ x^2-1>0 \]
\[ (x-1)(x+1)>0 \]
\[ x<-1 \quad \text{or} \quad x>1 \]
Intersecting with \(x\le4\),
\[ (-\infty,-1)\cup(1,4] \]
Therefore,
\[ \boxed{(-\infty,-1)\cup(1,4]} \]
\[ \boxed{\text{Correct Answer: (b)}} \]