If $f(x) = x^2 + 1$,then $fof(x)$ is equal to

Let $R$ be a relation '$ < $' from $A$ to $B$,where $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$ such that $(a, b) \in R \iff a < b$. Then $R \circ R^{-1}$ is:

Show that if $f: A \rightarrow B$ and $g: B \rightarrow C$ are one-one,then $g \circ f: A \rightarrow C$ is also one-one.

Let $f: \{2, 3, 4, 5\} \rightarrow \{3, 4, 5, 9\}$ and $g: \{3, 4, 5, 9\} \rightarrow \{7, 11, 15\}$ be functions defined as $f(2)=3, f(3)=4, f(4)=f(5)=5$ and $g(3)=g(4)=7$ and $g(5)=g(9)=11$. Find $g \circ f$.

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x)=\begin{cases} \log _e x & , x>0 \\ e^{-x} & , x \leq 0 \end{cases}$ and $g(x)=\begin{cases} x & , x \geq 0 \\ e^{x} & , x < 0 \end{cases}$. Then $gof: R \to R$ is . . . .

If $f(x)$ and $g(x)$ are two real valued functions such that $f(x)=3x-2$ and $g(x)=x^2+2$, then $[(g \circ f)+(f \circ g)](x) = $