Q. 54.0( 4 Votes )

If f(x) = loge(1 – x) and g(x) = [x], then determine each of the following functions:

i. f + g

ii. fg

iii.

iv.

Also, find (f + g)(–1), (fg)(0), and .

Answer :

Given f(x) = loge(1 – x) and g(x) = [x]


Clearly, f(x) takes real values only when 1 – x > 0


1 > x


x < 1


x (–∞, 1)


Thus, domain of f = (–∞, 1)


g(x) is defined for all real numbers x.


Thus, domain of g = R


i. f + g


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


(f + g)(x) = loge(1 – x) + [x]


Domain of f + g = Domain of f Domain of g


Domain of f + g = (–∞, 1) R


Domain of f + g = (–∞, 1)


Thus, f + g : (–∞, 1) R is given by (f + g)(x) = loge(1 – x) + [x]


ii. fg


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


(fg)(x) = loge(1 – x) × [x]


(fg)(x) = [x]loge(1 – x)


Domain of fg = Domain of f Domain of g


Domain of fg = (–∞, 1) R


Domain of fg = (–∞, 1)


Thus, f – g : (–∞, 1) R is given by (fg)(x) = [x]loge(1 – x)


iii.


We know



As earlier, domain of = (–∞, 1)


However, is defined for all real values of x (–∞, 1), except for the case when [x] = 0.


We have [x] = 0 when 0 ≤ x < 1 or x [0, 1)


When 0 ≤ x < 1, will be undefined as the division result will be indeterminate.


Domain of = (–∞, 1) – [0, 1)


Domain of = (–∞, 0)


Thus, : (–∞, 0) R is given by


iv.


We know



As earlier, domain of = (–∞, 1)


However, is defined for all real values of x (–∞, 1), except for the case when loge(1 – x) = 0.


loge(1 – x) = 0 1 – x = 1 or x = 0


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


Domain of = (–∞, 1) – {0}


Domain of = (–∞, 0) (0, ∞)


Thus, : (–∞, 0) (0, ∞) R is given by


We have (f + g)(x) = loge(1 – x) + [x], x (–∞, 1)


We need to find (f + g)(–1).


Substituting x = –1 in the above equation, we get


(f + g)(–1) = loge(1 – (–1)) + [–1]


(f + g)(–1) = loge(1 + 1) + (–1)


(f + g)(–1) = loge2 – 1


Thus, (f + g)(–1) = loge2 – 1


We have (fg)(x) = [x]loge(1 – x), x (–∞, 1)


We need to find (fg)(0).


Substituting x = 0 in the above equation, we get


(fg)(0) = [0]loge(1 – 0)


(fg)(0) = 0 × loge1


(fg)(0) = 0


Thus, (fg)(0) = 0


We have, x (–∞, 0)


We need to find


However, is not in the domain of.


Thus, does not exist.


We have, x (–∞, 0) (0, ∞)


We need to find


Substituting in the above equation, we get






Thus,


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