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If this is in the context of finding a root of an equation, the answer is to make some guesses. Find value x1 and x2 such that f(x1) and f(x2) have opposite signs. Then, provided that f is a continuous function over (x1, x2), the bisection method will find its root.

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Q: How do you find the The initial range of Bisection method?
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What is the defference between bisection method and newton method?

there are three variable are to find but in newton only one variable is taken at a time of a single iteration


To find the roots of 2xx-5x plus 1 equals 0 by bisection method?

I'm not familiar with the "bisection method" to find the roots of 2x2-5x+1 = 0 but by completing the square or using the quadratic equation formula you'll find that the solution is: x = (5 + or - the square root of 17) over 4 Hope that helps.


Why it is advantageous to combine Newton Raphson method and Bisection method to find the root of an algebraic equation of single variable?

An improved root finding scheme is to combine the bisection and Newton-Raphson methods. The bisection method guarantees a root (or singularity) and is used to limit the changes in position estimated by the Newton-Raphson method when the linear assumption is poor. However, Newton-Raphson steps are taken in the nearly linear regime to speed convergence. In other words, if we know that we have a root bracketed between our two bounding points, we first consider the Newton-Raphson step. If that would predict a next point that is outside of our bracketed range, then we do a bisection step instead by choosing the midpoint of the range to be the next point. We then evaluate the function at the next point and, depending on the sign of that evaluation, replace one of the bounding points with the new point. This keeps the root bracketed, while allowing us to benefit from the speed of Newton-Raphson.


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Using the projectile motion equations and given the initial velocity and angle, we can calculate the time the shell is in the air. Then, we can find the horizontal range by multiplying the time of flight by the horizontal component of the initial velocity. The horizontal range in this case is about 1056 meters.


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