Find f + g, f − g,

Question:

Find $1+g, 1-g, c f(c \in \mathrm{R}, c \neq 0), f g, \frac{1}{f}$ and $\frac{f}{g}$ in each of the following:

(a) If $f(x)=x^{3}+1$ and $g(x)=x+1$

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

Solution:

(a) Given:

$f(x)=x^{3}+1$ and $g(x)=x+1$

Thus,

$(f+g)(x): R \rightarrow R$ is given by $(f+g)(x)=f(x)+g(x)=x^{3}+1+x+1=x^{3}+x+2$

$(f-g)(x): R \rightarrow R$ is given by $(f-g)(x)=f(x)-g(x)=\left(x^{3}+1\right)-(x+1)=x^{3}+1-x-1=x^{3}-x$

$c f: R \rightarrow R$ is given by $(c f)(x)=c\left(x^{3}+1\right)$.

$(f g)(x): R \rightarrow R$ is given by $(f g)(x)=f(x) \cdot g(x)=\left(x^{3}+1\right)(x+1)=(x+1)\left(x^{2}-x+1\right)(x+1)=(x+1)^{2}\left(x^{2}-x+1\right)$

$\frac{1}{f}: R-\{-1\} \rightarrow R$ is given by $\left(\frac{1}{f}\right)(x)=\frac{1}{f(x)}=\frac{1}{\left(x^{3}+1\right)}$

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

Note that : $\left.\left(x^{3}+1\right)=(x+1)\left(x^{2}-x+1\right)\right]$

(b) Given:

$f(x)=\sqrt{x-1}$ and $g(x)=\sqrt{x+1}$

Thus,

$(f+g)):[1, \infty) \rightarrow R$ is defined by $(f+g)(x)=f(x)+g(x)=\sqrt{x-1}+\sqrt{x+1}$

$(f-g)):[1, \infty) \rightarrow R$ is defined by $(f-g)(x)=f(x)-g(x)=\sqrt{x-1}-\sqrt{x+1}$

$c f:[1, \infty) \rightarrow R$ is defined by $(c f)(x)=c \sqrt{x-1}$.

$(f g):[1, \infty) \rightarrow R$ is defined by $(f g)(x)=f(x) \cdot g(x)=\sqrt{x-1} \times \sqrt{x+1}=\sqrt{x^{2}-1} .$

$\frac{1}{f}:(1, \infty) \rightarrow R$ is defined by $\left(\frac{1}{\mathrm{f}}\right)(\mathrm{x})=\frac{1}{\mathrm{f}(\mathrm{x})}=\frac{1}{\sqrt{\mathrm{x}-1}}$

$\frac{\mathrm{f}}{\mathrm{g}}:[1, \infty) \rightarrow \mathrm{R}$ is defined by $\left(\frac{\mathrm{f}}{\mathrm{g}}\right)(\mathrm{x})=\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\frac{\sqrt{\mathrm{x}-1}}{\sqrt{\mathrm{x}+1}}=\sqrt{\frac{\mathrm{x}-1}{\mathrm{x}+1}}$.

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