# How to guess initial intervals for bisection method in order to reduce the no. of iterations?

SO, A function $f(x)$ is given to me and but the initial intervals are not given. I need to find the root of the equation using Bisection method.

Sometimes when I randomly guess the initial interval which obeys $f(a)·f(b)<0$ (where $a$ and $b$ are initial intervals) the no of iterations gets ridiculously large, sometimes 10-15 iterations. Since, I'll only have 3 hours in examination, I don't want to waste too much time on a single question. I know that the closer to the actual root my initial guesses are the lesser the no of iterations will be required.

Is there any quick way to calculate the root using calculator. I have a casio fx-991ES. I have tried using the solve equation function but for 2nd and 3rd degree equations calculator gives 2 and 3 roots which is what it is supposed to do. But since I'll be finding only one root with bisection method, I am confused on which root to take.

• Please describe exactly what you expect as task in your examination. If you were to compute some bisection steps, I would expect that you will be given an admissible initial interval. To not do so would horribly increase the correction effort. But then there certainly also exist masochistic examiners. – LutzL Jan 17 '16 at 11:55

For polynomials, you can use some outer root radius estimate and calculate a function value table to find intervals with changing signs. There is no guarantee to find any such interval, for instance for the polynomial $f(x)=(x-\sqrt2)^2-10^{-8}$, which has two real roots, it is highly improbable to land in the small interval of negative values.
The bisection method is inherently slow. To reduce an initial interval length by a factor of $2^n$ always requires exactly $n$ steps. If you want faster convergence with a bracketed method use the Illinois variant of regula falsi, it is only modestly more complex than the bisection method.