Find the Domain of \(f(x)=\sqrt{9-x}+\dfrac1{\sqrt{x^2-16}}\)

Solution

Given

\[ f(x)=\sqrt{9-x}+\frac1{\sqrt{x^2-16}} \]

For the function to be defined:

• The quantity inside each square root must be non-negative.
• The denominator must not be zero.

Condition 1

From

\[ \sqrt{9-x} \]

we must have

\[ 9-x\ge0 \] \[ x\le9 \]

Condition 2

Since

\[ \frac1{\sqrt{x^2-16}} \]

has a square root in the denominator, we need

\[ x^2-16>0 \] \[ x^2>16 \] \[ |x|>4 \]

Therefore,

\[ x<-4 \quad \text{or} \quad x>4 \]

Combine the Conditions

We now intersect:

\[ x\le9 \]

with

\[ x<-4 \quad \text{or} \quad x>4 \]

Hence the domain is

\[ (-\infty,-4)\cup(4,9] \]