Let f(x) = 1/√(16 - x² )

Clearly, f(x) is defined only when 16 - x² ≠ 0

Now, 16 - x² ≠ 0

=> x² - 16 ≠ 0

=> (x - 4)*(x + 4) ≠ 0

=> x ≠ -4, 4

So, domain of f(x) = R - {-4, 4}

Again let y = 1/√(16 - x² )

squaring on both side, we get

      y² = 1/(16 - x² )

=> 16 - x² = 1/y²

=> 16 - 1/y² = x² 

=> x = ± √(16 - 1/y² )

Clearly, x is defined, when y ≠ 0

So, range of f(x) = R - {0}