Q. 24.5( 2 Votes )

Let f(x) = 2x + 5 and g(x) = x2 + x. Describe

i. f + g

ii. f – g

iii. fg

iv.

Find the domain in each case.

Answer :

Given f(x) = 2x + 5 and g(x) = x2 + x


Clearly, 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) = 2x + 5 + x2 + x


(f + g)(x) = x2 + 3x + 5


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


The domain of (f + g) is R


ii. f – g


We know (f – g)(x) = f(x) – g(x)


(f – g)(x) = 2x + 5 – (x2 + x)


(f – g)(x) = 2x + 5 – x2 – x


(f – g)(x) = 5 + x – x2


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


The domain of (f – g) is R


iii. fg


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


(fg)(x) = (2x + 5)(x2 + x)


(fg)(x) = 2x(x2 + x) + 5(x2 + x)


(fg)(x) = 2x3 + 2x2 + 5x2 + 5x


(fg)(x) = 2x3 + 7x2 + 5x


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


The domain of fg is R


iv.


We know



Clearly, is defined for all real values of x, except for the case when x2 + x = 0.


x2 + x = 0


x(x + 1) = 0


x = 0 or x + 1 = 0


x = 0 or –1


When x = 0 or –1, will be undefined as the division result will be indeterminate.


Thus, domain of = R – {–1, 0}


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