## Precalculus (6th Edition) Blitzer

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 $A=4$.
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 $A=4$