Question

Find f + g, f − g, cf (c ∈ R, c ≠ 0), fg, $\frac{1}{f}\mathrm{and}\frac{f}{g}$in each of the following:

(a) If f(x) = x³ + 1 and g(x) = x + 1
(b) If $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\sqrt{x+1}$

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).
$\frac{1}{f}:R-\left\{-1\right\}\to R\text{is}\mathrm{given}\mathrm{by}\left(\frac{1}{f}\right)\left(x\right)=\frac{1}{f\left(x\right)}=\frac{1}{\left(x^3+1\right)}.$
$\frac{f}{g}:R-\left\{-1\right\}\to R\text{is}\mathrm{given}\mathrm{by}\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(x^3+1\right)}{\left(x+1\right)}=\frac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)}=\left(x^2-x+1\right).$

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

(b) Given:
$f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\sqrt{x+1}$
Thus,
(f + g) ) : [1, ∞) → R is defined by (f + g) (x) = f (x) + g (x) = $\sqrt{x-1}+\sqrt{x+1}$.
(f $-$ g) ) : [1, ∞) → R is defined by (f $-$ g) (x) = f (x) $-$ g (x) = $\sqrt{x-1}-\sqrt{x+1}$ .
cf : [1, ∞) → R is defined by (cf) (x) = $c\sqrt{x-1}$ .
(fg) : [1, ∞) → R is defined by (fg) (x) = f(x).g(x) = $\sqrt{x-1}\times \sqrt{x+1}=\sqrt{x^2-1}$ .
$\frac{1}{f}:\left(1,\infty \right)\to R\text{is}\mathrm{defined}\mathrm{by}\left(\frac{1}{\mathrm{f}}\right)\left(\mathrm{x}\right)=\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{1}{\sqrt{\mathrm{x}-1}}.$
$\frac{\mathrm{f}}{\mathrm{g}}:[1,\infty )\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{by}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)\left(\mathrm{x}\right)=\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{\sqrt{\mathrm{x}-1}}{\sqrt{\mathrm{x}+1}}=\sqrt{\frac{\mathrm{x}-1}{\mathrm{x}+1}}.$