Numerical approximation of function root using the secant method

Question

The task is as follows: Find the biggest negative root of the following equation using the secant method: $$\tan{x}-x+\frac{1}{2}=0$$ with $\epsilon =0.02$ being the max error.

So what I don't understand in this task is the meaning of the biggest negative root. When I graph the function, it has infinite negative roots due to $tan$ function. What is the meaning of biggest negative root?

Answer 1

$\tan x-x$ has derivative $\tan^2x$ and is thus monotonically increasing on every interval $I_k=(k\pi-\frac\pi2,k\pi+\frac\pi2)$ with a triple root at $x=0$ and a horizontal inflection point at $x=k\pi$ for $k\ne 0$. Thus there is exactly one solution to $\tan x-x=c$ in each of these intervals, and for $k=0$ the solution for $c=-\frac12$ in $I_0$ is negative, so that this is already the largest negative root.

If you want to solve the problem of finding the root in $I_k$ in a boring way without surprises, which often is the most robust way, you use your method to solve $$ x-k\pi=\arctan(x-\frac12). $$

By using the first Taylor terms one finds first guesses for the root in $I_0$ from $$ x+\frac13x^3-x+\frac12=0\iff x=-\sqrt[\Large3]{\frac32}=-1.14... $$