## Precalculus (6th Edition) Blitzer

The number for which the function f\left( x \right)=\left\{ \begin{align} & 7x\text{ if }x<6 \\ & \text{41 if }x=6 \\ & {{x}^{2}}+6\text{ if }x>6 \end{align} \right. is discontinuous at $x=6$
Consider the function f\left( x \right)=\left\{ \begin{align} & 7x\text{ if }x<6 \\ & \text{41 if }x=6 \\ & {{x}^{2}}+6\text{ if }x>6 \end{align} \right., Check the discontinuity of the function at $x=6$ Find the value of $f\left( x \right)$ at $x=6$, From the definition of the function, $f\left( 6 \right)=41$ The function is defined at the point $x=6$. Now find the value of $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)$, First find the left-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{6}^{-}}}{\mathop{\lim }}\,f\left( x \right)=7\left( 6 \right)=42$ Now find the right-hand limit of $\,f\left( x \right)$, That is, $\underset{x\to {{6}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{6}^{2}}+6=42$ Since the left-hand limit and right-hand limit are equal, that is $\underset{x\to {{6}^{-}}}{\mathop{\lim }}\,f\left( x \right)=42=\underset{x\to {{6}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=42$ From the above steps, $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 6 \right)$ Thus, the function does not satisfy the third property of being continuous. Hence, the function f\left( x \right)=\left\{ \begin{align} & 7x\text{ if }x<6 \\ & \text{41 if }x=6 \\ & {{x}^{2}}+6\text{ if }x>6 \end{align} \right. is discontinuous at $x=6$. Thus, the number for which the function f\left( x \right)=\left\{ \begin{align} & 7x\text{ if }x<6 \\ & \text{41 if }x=6 \\ & {{x}^{2}}+6\text{ if }x>6 \end{align} \right. is discontinuous at $x=6$.