Question

Find the range of the following functions given by f(x) = `3/(2 - x^2)`

Answer 1

Let f(x) = y

y = `3/(2 - x^2)`

⇒ 2 – x²

= `3/"y"`

⇒ = `2 - 3/y`

But, we know that, x² ≥ 0

`2 - 3/y ≥ 0`

⇒ `(2y - 3)/y ≥ 0`

⇒ y > 0 and 2y – 3 ≥ 0

⇒ y > 0 and 2y ≥ 3

⇒ y > 0 and y ≥ `3/2`

Or f(x) > 0 and f(x) ≥ `3/2`

 f(x) ∈ `( – oo, 0) ∪ (3/2 , oo)`

⇒ f(x) ∈ `( – oo, 0) ∪ (3/2 , oo)`

Therefore, the range of f = `(-oo, 0) ∪ (3/2 , oo)`.