91Ó°ÊÓ

The given composite function is:

$$\begin{gathered} f\left(x\right)=x^2+1 \\ g\left(x\right)=\sqrt{x-1} \end{gathered}$$

The domain of $f$ is the set of all real numbers andg is $\{x\overline{)xâ‰\yen 1\}}$.

$(f∘g)\left(x\right)=f\left(g\right(x\left)\right)$

Now substitute $g\left(x\right)=\sqrt{x-1}$in the function $f\left(g\right(x\left)\right)$.

Then the function will become $f\left(\sqrt{x-1}\right)$.

Now replace x with $\sqrt{x-1}$in $f\left(x\right)=x^2+1$,

$$\begin{gathered} f\left(\sqrt{x-1}\right)={\left(\sqrt{x-1}\right)}^2+1 \\ =x-1+1 \\ =x \end{gathered}$$

Therefore, the domain of$f∘g$is$\left\{x\overline{)xâ‰\yen 1}\right\}$and$(f∘g)\left(x\right)=x$.

$(g∘f)\left(x\right)=g\left(f\right(x\left)\right)$

Now substitute localid="1646231086303" $x^2+1$in the function $g\left(f\right(x\left)\right)$.

Then the function will become $g(x^2+1)$.

Now replace x with $x^2+1$in $g\left(x\right)=\sqrt{x-1}$,

$$\begin{gathered} g(x^2+1)=\sqrt{(x^2+1)-1} \\ =\sqrt{x^2+1-1} \\ =\left|x\right| \end{gathered}$$

Therefore, the domain of $g∘f$is the set of all real numbers and $g∘f\left(x\right)=\left|x\right|$.

c)$f∘f\left(x\right)=f\left(f\right(x\left)\right)$

Now substitute $f\left(x\right)=x^2+1$in the function $f\left(f\right(x\left)\right)$.

Then the function will become $f(x^2+1)$.

Now replace x with $x^2+1$in $f\left(x\right)=x^2+1$,

$$\begin{gathered} f(x^2+1)={(x^2+1)}^2+1=x^4+2x^2+1+1 \\ =x^4+2x^2+2 \end{gathered}$$

Therefore, the domain of $f∘f$is the set of all real numbers and$f∘f\left(x\right)=x^4+2x^2+2$.

$(g∘g)\left(x\right)=g\left(g\right(x\left)\right)$

Now substitute $g\left(x\right)=\sqrt{x-1}$in the function $g\left(g\right(x\left)\right)$.

Then the function will become $g\left(\sqrt{x-1}\right)$.

Now replace x with $\sqrt{x-1}$in $g\left(x\right)=\sqrt{x-1}$,

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

Solve the inequality to find x,

$$\begin{gathered} \sqrt{x-1}-1â‰\yen 0 \\ \sqrt{x-1}â‰\yen 1 \\ x-1â‰\yen 1 \\ xâ‰\yen 2 \end{gathered}$$

Therefore, the domain of $g∘g$is$\{x\overline{)xâ‰\yen 2\}}$and $g∘g\left(x\right)=\sqrt{\sqrt{x-1}-1}$.