Q. 1 A3.8( 5 Votes )

# Find f + g, f – g

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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