Q. 3 I4.1( 7 Votes )

# Find the domain and range of each of the following real valued functions:  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. Thus, range of f = Rate this question :

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