Answer
The statement “A piecewise function is always discontinuous at one or more numbers” is false.
Work Step by Step
Consider the piecewise function, $ f\left( x \right)=\left\{ \begin{align}
& x-1\text{ if }x\le 0 \\
& {{x}^{2}}+x-1\text{ if }x>0
\end{align} \right.$.
Since, $ x-1\text{ and }{{x}^{2}}+x-1$ both are continuous polynomial functions.
Therefore, the function can have the point of discontinuity only at $ a=0$.
Now check the discontinuity of the function at $ a=0$.
Find the value of $ f\left( x \right)$ at $ a=0$,
$\begin{align}
& f\left( 0 \right)=0-1 \\
& =-1
\end{align}$
The function is defined at the point $ a=0$.
Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$.
First check the left hand limit of $\,f\left( x \right)$.
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0-1=-1$
Now find the right hand limit of $\,f\left( x \right)$,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{0}^{2}}+0-1=-1$
Thus the left hand limit and right hand limit are equal, that is $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-1=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-1$
From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-1=f\left( 0 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& x-1\text{ if }x\le 0 \\
& {{x}^{2}}+x-1\text{ if }x>0
\end{align} \right.$ is continuous at $0$.
Thus, the function $ f\left( x \right)$ is not discontinuous for any number.
Hence, the statement is false.
