Find the domain and range of f(x) = 1/√(x2 - 16)

Question

Shiksha Nation · Accepted Answer

The expression under the square root must be positive.

x² - 16 > 0

⇒ (x - 4)(x + 4) > 0

⇒ x < -4 or x > 4

∴ Domain of f(x) = {x | x < -4 or x > 4}

For x < -4 or x > 4, denominator √(x² - 16) > 0

∴ Range of f(x) = (0, ¹/₀) → (0, ∞)

Hence, Range = (0, ∞)

Find the domain and range of f(x) = ¹/_{√(x² - 16)}