Question

Find the domain and range of each of the following real valued functions : f(x) = 1/√(16-x²)

f(x) = \(\frac{1}{\sqrt{16-x^2}}\)

Answer 1

f(x) = 1/√(16-x²)

We know,

The square of a real number is never negative. 

Clearly, 

f(x) takes real values only when 16 – x² ≥ 0 

⇒ 16 ≥ x² 

⇒ x² ≤ 16 

⇒ x² – 16 ≤ 0 

⇒ x² – 4² ≤ 0 

⇒ (x + 4)(x – 4) ≤ 0 

⇒ x ≥ –4 and x ≤ 4 

∴ x ∈ [–4, 4] 

In addition, 

f(x) is also undefined when 16 – x² = 0 because denominator will be zero and the result will be indeterminate. 

16 – x² = 0 

⇒ x = ±4 

Hence, 

x ∈ [–4, 4] – {–4, 4} 

∴ x ∈ (–4, 4) 

Thus, 

Domain of f = (–4, 4) 

Let f(x) = y

⇒ 1 = (16 – x²)y² 

⇒ 1 = 16y² – x²y² 

⇒ x²y² + 1 – 16y² = 0 

⇒ (y²)x² + (0)x + (1–16y²) = 0 

As x ∈ R,

The discriminant of this quadratic equation in x must be non-negative. 

⇒ 0² – 4(y²)(1–16y²) ≥ 0 

⇒ –4y²(1 – 16y²) ≥ 0 

⇒ 4y²(1 – 16y²) ≤ 0 

⇒ 1 – 16y² ≤ 0 

[∵ y² ≥ 0] 

⇒ 16y² – 1 ≥ 0 

⇒ (4y)² – 1² ≥ 0 

⇒ (4y + 1)(4y – 1) ≥ 0 

⇒ 4y ≤ –1 and 4y ≥ 1

However,

y is always positive because it is the reciprocal of a non-zero square root.