Question

Let $f\left(x\right)=\sqrt{x}andg\left(x\right)=x$ be two functions defined in the domain $R^{+}\cup \left\{0\right\}$. Find

(i) $(f+g)\left(x\right)$

(ii) $(f-g)\left(x\right)$

(iii) $\left(fg\right)\left(x\right)$

(iv) $\left(\frac{f}{g}\right)\left(x\right)$

Answer 1

(i) Using the algebraic operations on functions

Given: $f\left(x\right)=\sqrt{x}andg\left(x\right)=x$, defined in the domain $R^{+}\cup \left\{0\right\}$.

$(f+g)\left(x\right)=\sqrt{x}+x$

$(\because (f+g\left)\right(x)=f(x)+g(x\left)\right)$

Hence, $(f+g)\left(x\right)=\sqrt{x}+x$

(ii) Using the algebraic operations on functions

Given: $f\left(x\right)=\sqrt{x}andg\left(x\right)=x$, defined in the domain $R^{+}\cup \left\{0\right\}$.

$(f-g)\left(x\right)=\sqrt{x}-x$

$(\because (f-g\left)\right(x)=f(x)-g(x\left)\right)$

Hence, $(f-g)\left(x\right)=\sqrt{x}-x$

(iii) Using the algebraic operations on functions

Given: $f\left(x\right)=\sqrt{x}andg\left(x\right)=x$, defined in the domain $R^{+}\cup \left\{0\right\}$.

$\left(fg\right)\left(x\right)=x\sqrt{x}=x^{\frac{3}{2}}$

$(\because (fg\left)\right(x)=f(x).g(x\left)\right)$

Hence, $\left(fg\right)\left(x\right)=x^{\frac{3}{2}}$

(iv) Using the algebraic operations on functions

Given: $f\left(x\right)=\sqrt{x}andg\left(x\right)=x$, defined in the domain $R^{+}\cup \left\{0\right\}$.

$\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x}}{x}=x^{-\frac{1}{2}}$

$(\because \left(\frac{f}{g}\right)(x)=\frac{f\left(x\right)}{g\left(x\right)}.g(x)\ne 0)$

Hence, $\left(\frac{f}{g}\right)\left(x\right)=\frac{1}{\sqrt{x}},x\ne 0$