Let $\mathrm{f:} \mathrm{R} \rightarrow \mathrm{R}: \mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+1$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}: \mathrm{g}(\mathrm{x})=(\mathrm{x}+1)$.
Find:
(i) $(f+g)(x)$
(ii) $(f-g)(x)$
(iii) (1/f) (x)
(iv) $(f / g)(x)$
(i) Given:
$f(x)=x^{3}+1$ and $g(x)=x+1$
(i) To find: $(f+g)(x)$
$(f+g)(x)=f(x)+g(x)$
$=\left(x^{3}+1\right)+(x+1)$
$=x^{3}+1+x+1$
$=x^{3}+x+2$
Therefore,
$(f+g)(x)=x^{3}+x+2$
(ii) To find: $(f-g)(x)$
$(f-g)(x)=f(x)-g(x)$
$=\left(x^{3}+1\right)-(x+1)$
$=x^{3}+1-x-1$
$=x^{3}-x$
Therefore
$(f-g)(x)=x^{3}-x$
(iii) To find $:\left(\frac{1}{f}\right)(x)$
$\left(\frac{1}{f}\right)(x)=\left(\frac{1}{f(x)}\right)$
$=\left(\frac{1}{x^{3}+1}\right)$
Therefore,
$\left(\frac{1}{f}\right)(x)=\left(\frac{1}{x^{3}+1}\right)$
(iv) To find $:\left(\frac{f}{g}\right)(x)$
$\left(\frac{f}{g}\right)(x)=\left(\frac{f(x)}{g(x)}\right)$
$=\left(\frac{x^{3}+1}{x+1}\right)$
$=\left(\frac{x^{3}+1^{3}}{x+1}\right)$
$=\left(\frac{(x+1)\left(x^{2}-x+1\right)}{x+1}\right)$ (Because $\left.\mathrm{a}^{3}+\mathrm{b}^{3}=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{2}-\mathrm{ab}+\mathrm{b}^{2}\right)\right)$
Therefore,
$\left(\frac{f}{g}\right)(x)=x^{2}-x+1$
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