Consider the algorithm below, which prints the roots of the cubic polynomial  \(f(x)=x^3-2 x^2-9 x+18\).

\begin{array} {l}
\rule{0pt}{2.5ex} \textbf{ define} \ \text{ f (x) }  \\
\rule{0pt}{2.5ex} \quad \quad \textbf{return} \ \text{(x} ^3 - 2  \text{x}^2 - 9 \text{x} + 18) \\
\rule{0pt}{2.5ex}  \text{c} \leftarrow \text{f} \ (0) \\
\rule{0pt}{2.5ex} \textbf{if}\  \ \text{c < 0} \ \textbf{then}\\
\rule{0pt}{2.5ex} \quad \quad \text{c} \ \leftarrow \ \text{(- c)} \\
\rule{0pt}{2.5ex} \textbf{end if} \\
\rule{0pt}{2.5ex} \textbf{while} \ \text{ c > 0 } \\
\rule{0pt}{2.5ex} \quad \quad \textbf{if} \ \ \text{f (c) = 0 } \ \textbf{then}  \\
\rule{0pt}{2.5ex} \quad \quad \quad \quad \textbf{print} \ \text{c }   \\
\rule{0pt}{2.5ex} \quad \quad  \textbf{end if} \\
\rule{0pt}{2.5ex} \quad \quad  \textbf{if} \ \ \text{f (-c) = 0} \ \textbf{then} \\
\rule{0pt}{2.5ex} \quad \quad \quad \quad \textbf{print} \ \text{-c }   \\
\rule{0pt}{2.5ex} \quad \quad  \textbf{end if} \\
\rule{0pt}{2.5ex} \quad \quad  \text{c} \ \leftarrow \ \text{c - 1} \\
\rule{0pt}{2.5ex} \textbf{ end while } \\
\end{array}

In order, the algorithm prints the values

1. \(-3, 3, 2\)
2. \(-3, 2, 3\)
3. \(3, 2, -3\)
4. \(3, -3, 2\)

Consider the algorithm below, which uses the bisection method to estimate the solution to an equation in the form \(f(x)=0\).

The algorithm is implemented as follows.
 

Which value would be returned when the algorithm is implemented as given?

1. -0.351
2. -0.108
3. 3.25
4. 3.5
5. 4