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 .

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