Question

Find f + g, f - g, $cf(cϵR,c\ne 0)$. fg. $\frac{1}{f}$ and $\frac{f}{g}$ in each of the following :
(i) $f\left(x\right)=x^3+1$ and $g\left(x\right)=x+1$
(ii) $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\sqrt{x+1}$

Answer 1

We have,
$f\left(x\right)=x^3+1$ and $g\left(x\right)=x+1$
Now,
$f+g:R\to R$ given by $(f+g)\left(x\right)=x^3+x+2$
$f-g:R\to R$ given by $(f-g)\left(x\right)=x^3+1-(x+1)$
$=x^3-x$
$cf:R\to R$ given by $\left(cf\right)\left(x\right)=c(x^3+1)$
$fg:R\to R$ given by $\left(fg\right)\left(x\right)=(x^3+1)(x+1)$
$=x^4+x^3+x+1$
$\frac{1}{f}:R-\{-1\}\to R$ given by $\left(\frac{1}{f}\right)\left(x\right)=\frac{1}{x^3+1}$
$\frac{f}{g}:R-\{-1\}\to R$ given by $\left(\frac{f}{g}\right)\left(x\right)=\frac{(x+1)(x^2-x+1)}{x+1}$
$=x^2-x+1$

(ii) We have,
$f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\sqrt{x+1}$
Now,
$f+g:(1,\infty )\to R$ defined by $(f+g)=\left(x\right)=\sqrt{x-1}+\sqrt{x+1}$
$f-g:(1,\infty )\to R$ defined by $(f-g)=\left(x\right)=\sqrt{x-1}-\sqrt{x+1}$
$cf:(1,\infty )\to R$ defined by $\left(cf\right)\left(x\right)=c\sqrt{x-1}$
$fg:(1,\infty )\to R$ defined by $\left(fg\right)\left(x\right)=\left(\sqrt{x-1}\right)\left(\sqrt{x+1}\right)=\sqrt{x^2-1}$
$\frac{1}{f}:(1,\infty )\to R$ defined by $\left(\frac{1}{f}\right)\left(x\right)=\frac{1}{\sqrt{x-1}}$
$\frac{f}{g}:(1,\infty )\to R$ defined by $\left(\frac{f}{g}\right)\left(x\right)=\sqrt{\frac{x-1}{x+1}}$