Q. 1 A3.8( 5 Votes )

Find f + g, f – g

Answer :

i. f(x) = x3 + 1 and g(x) = x + 1


We have f(x) : R R and g(x) : R R


(a) f + g


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


(f + g)(x) = x3 + 1 + x + 1


(f + g)(x) = x3 + x + 2


Clearly, (f + g)(x) : R R


Thus, f + g : R R is given by (f + g)(x) = x3 + x + 2


(b) f – g


We know (f – g)(x) = f(x) – g(x)


(f – g)(x) = x3 + 1 – (x + 1)


(f – g)(x) = x3 + 1 – x – 1


(f – g)(x) = x3 – x


Clearly, (f – g)(x) : R R


Thus, f – g : R R is given by (f – g)(x) = x3 – x


(c) cf (c R, c ≠ 0)


We know (cf)(x) = c × f(x)


(cf)(x) = c(x3 + 1)


(cf)(x) = cx3 + c


Clearly, (cf)(x) : R R


Thus, cf : R R is given by (cf)(x) = cx3 + c


(d) fg


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


(fg)(x) = (x3 + 1)(x + 1)


(fg)(x) = (x + 1)(x2 – x + 1)(x + 1)


(fg)(x) = (x + 1)2(x2 – x + 1)


Clearly, (fg)(x) : R R


Thus, fg : R R is given by (fg)(x) = (x + 1)2(x2 – x + 1)


(e)


We know



Observe that is undefined when f(x) = 0 or when x = – 1.


Thus, : R – {–1} R is given by


(f)


We know



Observe that is undefined when g(x) = 0 or when x = –1.


Using x3 + 1 = (x + 1)(x2 – x + 1), we have




Thus, : R – {–1} R is given by


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