Question

Let f(x) = x² and g(x) = 2x+ 1 be two real functions. Find (f + g) (x), (f − g) (x), (fg) (x) and  \[\left( \frac{f}{g} \right) \left( x \right)\] .

 

Answer 1

Given:
 f (x)  = x² and g (x) = 2x + 1
Clearly, D (f) =  R and D (g) = R

\[\therefore D\left( f \pm g \right) = D\left( f \right) \cap D\left( g \right) = R \cap R = R\]

\[D\left( fg \right) = D\left( f \right) \cap D\left( g \right) = R \cap R = R\]

\[D\left( \frac{f}{g} \right) = D\left( f \right) \cap D\left( g \right) - \left\{ x: g\left( x \right) = 0 \right\} = R \cap R - \left\{ - \frac{1}{2} \right\} = R - \left\{ - \frac{1}{2} \right\}\]

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

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