# Let f :

Question:

Let $t:[0, \infty) \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=\sqrt{x}$ and $g(x)=x$. Find $f+g, f-g, f g$ and $\frac{f}{g}$.

Solution:

It is given that $f:[0, \infty) \rightarrow R$ and $g: R \rightarrow R$ such that $f(x)=\sqrt{x}$ and $g(x)=x$.

$D(f+g)=[0, \infty) \cap R=[0, \infty)$

So, $f+g:[0, \infty) \rightarrow R$ is given by

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

$D(f-g)=D(f) \cap D(g)=[0, \infty) \cap R=[0, \infty)$

So, f - g : [0, ∞) → R is given by

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

$D(f g)=D(f) \cap D(g)=[0, \infty) \cap R=[0, \infty)$

So, $f g:[0, \infty) \rightarrow R$ is given by

$(f g)(x)=f(x) g(x)=\sqrt{x} \cdot x=x^{3 / 2}$

$D\left(\frac{f}{g}\right)=[D(f) \cap D(g)-\{x: g(x)=0\}]=(0, \infty)$

So, $\frac{f}{g}:(0, \infty) \rightarrow R$ is given by

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