1.

Find the domain and range of each of the following real valued functions : f(x) = √(9-x2)

Answer»

f(x) = √(9-x2)

We know,

The square of a real number is never negative. 

Clearly, 

f(x) takes real values only when 9 – x2 ≥ 0 

⇒ 9 ≥ x2 

⇒ x2 ≤ 9 

⇒ x2 – 9 ≤ 0 

⇒ x2 – 32 ≤ 0 

⇒ (x + 3)(x – 3) ≤ 0 

⇒ x ≥ –3 and x ≤ 3 

∴ x ∈ [–3, 3] 

Thus,

Domain of f = [–3, 3] 

When x ∈ [–3, 3], we have 

0 ≤ 9 – x2 ≤ 9 

Hence,

0 ≤ \(\sqrt{9-x^2}\) ≤ 3

⇒ 0 ≤ f(x) ≤ 3

∴ f(x) ∈ [0, 3] 

Thus, 

Range of f = [0, 3]



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