This is not a homework problem. It is from a closed MITs 2016 advanced algorithms class.
Here is my proposed solution:
We can run the linear program
$$\begin{array}{ll} \text{maximize} & f^T x\\ \text{subject to} & A x \geq b\\ & f^T x \geq 1\\ & -8 \leq c^T x \leq 8\end{array}$$
If the linear program is infeasible then the problem is infeasible, and if the linear program is unbounded then the solution is $0$. Suppose neither of these is the case. Then $ -8\leq \frac{c^Tx}{f^Tx} \leq 8$, and the idea is to do binary search until the possible interval that $\frac{c^Tx}{f^Tx}$ lies in is of size $\leq \epsilon$.
So for the $i$th iteration of the algorithm suppose $ L \leq \frac{c^Tx}{f^Tx} \leq U$. $\frac{c^Tx}{f^Tx} \leq \frac{L+U}{2}$ if and only if $c^Tx - \frac{(f^Tx)(L+U)}{2} \leq 0$, since $ f^Tx \geq 1$. So if we run the linear program,
$$\begin{array}{ll} \text{minimize} & c^Tx - \frac{(f^Tx)(L+U)}{2} \\ \text{subject to} & A x \geq b\\ & f^T x \geq 1\\ & -8 \leq c^T x \leq 8\end{array}$$
, and we get a value of $0$ or less, then we have $ L \leq \frac{c^Tx}{f^Tx} \leq \frac{U + L}{2}$, else we have $ \frac{L + U}{2} \leq \frac{c^Tx}{f^Tx} \leq U$. We then repeat this process until $U-L \leq \epsilon$ and we take $U$ as our answer.
The running time of this algorithm is $k$, where $\frac{16}{2^k} \leq \epsilon$ so $k\leq \log(\frac{16}{\epsilon})$.
Is this a correct solution to the problem?
Edit: There is actually a $O(\frac{1}{\epsilon})$ solution(by someone else); However I'm still not sure if my $O(log(\frac{1}{\epsilon}))$ solution is any good.
Hint: Introduce optimization variable $y$, make the objective $y$ and append the constraint $$\frac{c^Tx}{f^Tx} \leq y$$ Note that the denominator is positive. Then, pick, say, $y =2$ and use linear programming to determine whether the constraints define an empty polytope or not. If the polytope is non-empty, then pick, say, $y = 1$ and so on. Once you find an empty polytope, use binary search to adjust $y$ till you find a "decent" approximation for the minimum. My guess is that this is what the authors meant by "within any degree of accuracy".