Q. 3 I4.1( 7 Votes )

Find the domain and range of each of the following real valued functions:


Answer :


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 =


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