Question

Find f + g, f − g, cf (c ∈ R, c ≠ 0), fg, \[\frac{1}{f}\text{  and } \frac{f}{g}\] in :

(a) If f(x) = x³ + 1 and g(x) = x + 1

Answer 1

(a) Given:
f (x)  = x³  + 1 and g (x) = x + 1
Thus,
(f + g) (x) : R → R is given by (f + g) (x) = f (x) + g (x) = x³ + 1 + x + 1 = x³ + x + 2.
(f - g) (x) : R → R is given by (f - g) (x) = f (x) - g (x) = (x³ + 1) - (x + 1 ) = x³ + 1 -x -  1 = x³ - x.
cf : R → R is given by (cf) (x) = c (x³  + 1).
(fg) (x) : R → R is given by (fg) (x) = f(x).g(x) = (x³ + 1) (x + 1) = (x + 1) (x² - x + 1) (x + 1) = (x + 1)² (x² - x + 1).

\[\frac{1}{f}: R - \left\{ - 1 \right\} \to R\text{  is given by} \left( \frac{1}{f} \right)\left( x \right) = \frac{1}{f\left( x \right)} = \frac{1}{\left( x^3 + 1 \right)} .\]

\[\frac{f}{g}: R - \left\{ - 1 \right\} \to \text{ R isgiven by} \left( \frac{f}{g} \right)\left( x \right) = \frac{f\left( x \right)}{g\left( x \right)} = \frac{\left( x^3 + 1 \right)}{\left( x + 1 \right)} = \frac{\left( x + 1 \right)\left( x^2 - x + 1 \right)}{\left( x + 1 \right)} = \left( x^2 - x + 1 \right) .\]

Note that : (x³ + 1) = (x + 1) (x² - x + 1)]