Answer
$f(x)=\begin{cases}
\dfrac{1}{x-2}+1,\text{for }x\in[0,2)\\
2,\text{for }x=2\\
\dfrac{x}{x-2}-3,\text{for }x\in(2,4]
\end{cases}$
Work Step by Step
Consider the function:
$f(x)=\begin{cases}
\dfrac{1}{x-2}+1,\text{for }x\in[0,2)\\
2,\text{for }x=2\\
\dfrac{x}{x-2}-3,\text{for }x\in(2,4]
\end{cases}$
$\displaystyle\lim_{x\rightarrow 2^{-}} f(x)=-\infty$
$\displaystyle\lim_{x\rightarrow 2^{+}} f(x)=\infty$
Therefore $f$ is not continuous in $x=2$.
$f(0.5)f(3.5)=(-1)(2.1)<0$ and there is a zero between $0.5$ and $3.5$.
