Answer
The value of A so that the function $ f\left( x \right)=\left\{ \begin{align}
& {{x}^{2}}\text{ if }x<1 \\
& Ax-3\text{ if }x\ge 1
\end{align} \right.$ is continuous at $1$ is $ A=4$.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& {{x}^{2}}\text{ if }x<1 \\
& Ax-3\text{ if }x\ge 1
\end{align} \right.$.
Find the value of $ f\left( x \right)$ at $ a=1$.
From the definition of the function, for $ x=1$, $ f\left( x \right)=Ax-3$ .
Then the value of $ f\left( x \right)$ at $ a=1$ is,
$\begin{align}
& f\left( 1 \right)=A\left( 1 \right)-3 \\
& =A-3
\end{align}$
Now find the value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$,
First find the left hand limit of $\,f\left( x \right)$,
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)={{\left( 1 \right)}^{2}}=1$
Now find the right hand limit of $\,f\left( x \right)$,
$\begin{align}
& \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=A\left( 1 \right)-3 \\
& =A-3
\end{align}$
For the function to be continuous at the point $ a=1$, the $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ must exist.
The value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ can exist only if $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
From the above steps, solve the equation $1=A-3$ for the value of A.
$\begin{align}
& 1=A-3 \\
& A=3+1 \\
& A=4
\end{align}$
Therefore, the value of A so that the function $ f\left( x \right)=\left\{ \begin{align}
& {{x}^{2}}\text{ if }x<1 \\
& Ax-3\text{ if }x\ge 1
\end{align} \right.$ is continuous at $1$ is $ A=4$
