Let $D$ be the domain of the real valued function $f$ defined by $f(x)=\sqrt{25-x^2}$. Then, write $D$.
Given function is, $f(x)=\sqrt{25-x^2}$
$$\begin{aligned} & \text { For real valued of } f(x) \quad 25-x^2 \geq 0 \\ & x^2 \leq 25 \\ & -5 \leq x \leq+5 \\ \therefore\quad& D=[-5,5] \end{aligned}$$
If $f, g: R \rightarrow R$ be defined by $f(x)=2 x+1$ and $g(x)=x^2-2, \forall x \in R$, respectively. Then, find $gof$.
Given that,
$$\begin{aligned} f(x)=2 x+1 \text { and } g(x) & =x^2-2, \forall x \in R \\ \therefore\quad g \circ f & =g\{f(x)\} \\ & =g(2 x+1)=(2 x+1)^2-2 \\ & =4 x^2+4 x+1-2 \\ & =4 x^2+4 x-1 \end{aligned}$$
Let $f: R \rightarrow R$ be the function defined by $f(x)=2 x-3, \forall x \in R$. Write $f^{-1}$.
$$\begin{aligned} \text{Given that,} \quad f(x) & =2 x-3, \forall x \in R \\ \text{Now, let }\quad y & =2 x-3 \\ 2 x & =y+3 \\ x & =\frac{y+3}{2} \\ \therefore\quad f^{-1}(x) & =\frac{x+3}{2} \end{aligned}$$
If $A=\{a, b, c, d\}$ and the function $f=\{(a, b),(b, d),(c, a),(d, c)\}$, write $f^{-1}$.
$$\begin{aligned} \text{Given that,}\quad A & =\{a, b, c, d\} \\ \text{and }\quad f & =\{(a, b),(b, d),(c, a),(d, c)\} \\ f^{-1} & =\{(b, a),(d, b),(a, c),(c, d)\} \end{aligned} $$
If $f: R \rightarrow R$ is defined by $f(x)=x^2-3 x+2$, write $f\{f(x)\}$.
$$\begin{aligned} \text{Given that,}\quad f(x) & =x^2-3 x+2 \\ \therefore\quad f\{f(x)\} & =f\left(x^2-3 x+2\right) \\ & =\left(x^2-3 x+2\right)^2-3\left(x^2-3 x+2\right)+2 \\ & =x^4+9 x^2+4-6 x^3-12 x+4 x^2-3 x^2+9 x-6+2 \\ & =x^4+10 x^2-6 x^3-3 x \\ f\{f(x)\} & =x^4-6 x^3+10 x^2-3 x \end{aligned}$$