Q. 105.0( 1 Vote )

# Let f(x) = x^{2} and g(x) = 2x + 1 be two real functions. Find (f + g)(x), (f – g)(x), (fg)(x) and .

Answer :

Given f(x) = x^{2} and g(x) = 2x + 1

Both f(x) and g(x) are defined for all x ∈ R.

Hence, domain of f = domain of g = R

i. f + g

We know (f + g)(x) = f(x) + g(x)

⇒ (f + g)(x) = x^{2} + 2x + 1

∴ (f + g)(x) = (x + 1)^{2}

Clearly, (f + g)(x) is defined for all real numbers x.

∴ Domain of (f + g) is R

Thus, f + g : R → R is given by (f + g)(x) = (x + 1)^{2}

ii. f – g

We know (f – g)(x) = f(x) – g(x)

⇒ (f – g)(x) = x^{2} – (2x + 1)

∴ (f – g)(x) = x^{2} – 2x – 1

Clearly, (f – g)(x) is defined for all real numbers x.

∴ Domain of (f – g) is R

Thus, f – g : R → R is given by (f – g)(x) = x^{2} – 2x – 1

iii. fg

We know (fg)(x) = f(x)g(x)

⇒ (fg)(x) = x^{2}(2x + 1)

∴ (fg)(x) = 2x^{3} + x^{2}

Clearly, (fg)(x) is defined for all real numbers x.

∴ Domain of fg is R

Thus, fg : R → R is given by (fg)(x) = 2x^{3} + x^{2}

iv.

We know

Clearly, is defined for all real values of x, except for the case when 2x + 1 = 0.

2x + 1 = 0

⇒ 2x = –1

When, will be undefined as the division result will be indeterminate.

Thus, the domain of = R –

Rate this question :

Let f(x) = 2x + 5 and g(x) = x^{2} + x. Describe

i. f + g

ii. f – g

iii. fg

iv.

Find the domain in each case.

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