Answer
$f(x)=\begin{cases}
\dfrac{1}{x-2},\text{for }x\in[0,2)\cup(2,4]\\
2,\text{for }x=2
\end{cases}$
Work Step by Step
Consider the function:
$f(x)=\begin{cases}
\dfrac{1}{x-2},\text{for }x\in[0,2)\cup(2,4]\\
2,\text{for }x=2
\end{cases}$
$\displaystyle\lim_{x\rightarrow 2^{-}} f(x)=-\infty$
$\displaystyle\lim_{x\rightarrow 2^{+}} f(x)=\infty$
Therefore $f$ is not continuous in $x=2$.
$f(1)f(3)=(-1)(1)<0$, but there is no zero between $1$ and $3$.
