1.

Find the domain of each of the following real valued functions of real variable :i. f(x) = 1/xii. f(x) = \(\frac{1}{x-7}\)iii. f(x) = \(\frac{3x-2}{x+1}\)iv. f(x) = \(\frac{2x+1}{x^2+9}\) v. f(x) = \(\frac{x^2+2x+1}{x^2-8x+12}\)

Answer»

i. f(x) = 1/x

Clearly,

f(x) is defined for all real values of x, 

Except for the case when x = 0. 

When x = 0, 

f(x) will be undefined as the division result will be indeterminate. 

Thus, 

Domain of f = R – {0}

ii. f(x) = \(\frac{1}{x-7}\) 

Clearly, 

f(x) is defined for all real values of x, 

Except for the case when x – 7 = 0 or x = 7. 

When x = 7,

f(x) will be undefined as the division result will be indeterminate. 

Thus,

Domain of f = R – {7}

iii. f(x) = \(\frac{3x-2}{x+1}\)

Clearly, 

f(x) is defined for all real values of x, 

Except for the case when x + 1 = 0 or x = –1. 

When x = –1, 

f(x) will be undefined as the division result will be indeterminate. 

Thus, 

Domain of f = R – {–1}

iv. f(x) = \(\frac{2x+1}{x^2+9}\) 

Clearly,

f(x) is defined for all real values of x, 

Except for the case when x2 – 9 = 0. 

x2 – 9 = 0 

⇒ x2 – 32 = 0 

⇒ (x + 3)(x – 3) = 0 

⇒ x + 3 = 0 or x – 3 = 0 

⇒ x = ±3 

When x = ±3, 

f(x) will be undefined as the division result will be indeterminate. 

Thus, 

Domain of f = R – {–3, 3}

v. f(x) = \(\frac{x^2+2x+1}{x^2-8x+12}\)  

Clearly, 

f(x) is defined for all real values of x, 

except for the case when x2 – 8x + 12 = 0. 

x2 – 8x + 12 = 0 

⇒ x2 – 2x – 6x + 12 = 0 

⇒ x(x – 2) – 6(x – 2) = 0 

⇒ (x – 2)(x – 6) = 0 

⇒ x – 2 = 0 or x – 6 = 0 

⇒ x = 2 or 6 

When x = 2 or 6, 

f(x) will be undefined as the division result will be indeterminate. 

Thus, 

Domain of f = R – {2, 6}



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