Question

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

(i) f (x) = (ax + b)/(bx - a)

(ii) f (x) = (ax - b)/(cx - d)

(iii) f (x) = √(x - 1)

(iv) f (x) = √(x - 3)

(v) f (x) = (x - 2)/(2 - x)

(vi) f (x) = |x - 1|

(vii) f (x) = -|x|

(viii) f (x) = √(9 - x²)

Answer 1

(i) Since, f(x) is defined for all real values of x, except for the case when bx – a = 0 or x = a/b.

Domain (f) = R – (a/b)

Suppose f (x) = y

(ax + b)/(bx - a) = y

ax + b = y(bx – a)

ax + b = bxy – ay

ax – bxy = –ay – b

x(a – by) = –(ay + b)

∴ x = – (ay + b)/(a - by)

When a – by = 0 or y = a/b

Thus, f(x) cannot take the value a/b.

∴ Range (f) = R – (a/b)

(ii) Since, f(x) is defined for all real values of x, except for the case when cx – d = 0 or x = d/c. Domain (f) = R – (d/c) 

Suppose f (x) = y

(ax - b)/(cx - d) = y

ax – b = y(cx – d)

ax – b = cxy – dy

ax – cxy = b – dy

x(a – cy) = b – dy

∴ x = (b - dy)/(a - cy)

When a – cy = 0 or y = a/c,

Thus, f(x) cannot take the value a/c.

∴ Range (f) = R – (a/c)

(iii) f (x) = √(x - 1)

As we know the square of a real number is never negative.

Now, f(x) takes real values only when x – 1 ≥ 0

x ≥ 1

∴ x ∈ [1, ∞)

Hence, domain (f) = [1, ∞)

When x ≥ 1, we have x – 1 ≥ 0

Thus, √(x-1) ≥ 0 ⇒ f (x) ≥ 0  

f(x) ∈ [0, ∞)

∴ Range (f) = [0, ∞)

(iv) As we know the square of a real number is never negative.

Now, f (x) takes real values only when x – 3 ≥ 0

x ≥ 3

∴ x ∈ [3, ∞)

Domain (f) = [3, ∞)

When x ≥ 3, we have x – 3 ≥ 0

Thus, √(x-3) ≥ 0 ⇒ f (x) ≥ 0

f(x) ∈ [0, ∞)

∴ Range (f) = [0, ∞)

(v) Since, f(x) is defined for all real values of x, except for the case when 2 – x = 0 or x = 2.

Domain (f) = R – {2}

Now, we have, f (x) = (x - 2)/(2 - x) 

f (x) = -(2 - x)/(2 - x)

= –1

When x ≠ 2, f(x) = –1

∴ Range (f) = {–1}

(vi) As we know

Thus, f(x) is defined for all real numbers x.

Domain (f) = R

When, x < 1, we have x – 1 < 0 or 1 – x > 0.

|x – 1| > 0 ⇒ f(x) > 0

When, x ≥ 1, we have x – 1 ≥ 0.

|x – 1| ≥ 0 ⇒ f(x) ≥ 0

∴ f(x) ≥ 0 or f(x) ∈ [0, ∞)

Range (f) = [0, ∞)

(vii) As we know

Thus, f(x) is defined for all real numbers x.

Domain (f) = R

When, x < 0, we have –|x| < 0

f (x) < 0

When, x ≥ 0, we have –x ≤ 0.

–|x| ≤ 0 ⇒ f (x) ≤ 0

∴ f (x) ≤ 0 or f (x) ∈ (–∞, 0]

Range (f) = (–∞, 0]

(viii) As we know the square of a real number is never negative.

Now, 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]

Domain (f) = [–3, 3]

When, x ∈ [–3, 3], we have 0 ≤ 9 – x² ≤ 9

0 ≤ √(9 - x²) ≤ 3 ⇒ 0 ≤ f (x) ≤ 3

 ∴ f(x) ∈ [0, 3]

The of range (f) = [0, 3]