Question

If $f\left(x\right)=\sqrt{x+3}$ and $g\left(x\right)=x^2+1$ be two real functions, then find $f\circ g$ and $g\circ f$.

Answer 1

$f\left(x\right)=\sqrt{x+3}$

For domain, $x+3\ge 0$

$\Rightarrow$ $x\ge -3$

Domain of $f=[-3,\infty )$

Range of $f=(-3,\infty )$

Similarly, range of $g=(1,\infty )$

Then, rangle of f is subset of domain $g$ and range of $g$ is subset of $f$.

$\therefore$ $f\circ g$ and $g\circ f$ exist.

$\Rightarrow$ $f\circ g\left(x\right)=f\left[g\right(x\left)\right]$

$=f(x^2+1)$ ......... [ Since, $g(x{)}^2+1$ ]

$=\sqrt{x^2+1+3}$ ......... [ Since, $f\left(x\right)=\sqrt{x+3}$ ]

$=\sqrt{x^2+4}$

$\Rightarrow$ $g\circ f\left(x\right)=g\left[f\right(x\left)\right]$

$=g\left(\sqrt{x+3}\right)$......... [ Since, $f\left(x\right)=\sqrt{x+3}$ ]

$=(\sqrt{x+3}{)}^2+1$......... [ Since, $g(x{)}^2+1$ ]

$=x+3+1$

$=x+4$