9514 1404 393
Answer:
x ≈ 2.367796
Explanation:
There are a number of ways to use "successive approximation" to find a solution to an equation. You have not specified the method or the required accuracy.
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One of the more tedious is the bisection method, which computes the value of the function at the midpoint of an interval, to see if that midpoint should be a new interval bound. This method requires about 3 iterations per decimal place of the answer.
Another method is the secant method. It computes the x-intercept of the secant line joining the points at the ends of the interval containing the solution. The function value at this x-intercept is used to see if that point should be a new interval bound. This method converges slightly faster than the bisection method. It is speeded by adjusting both ends of the interval.
Newton's method iteration converges at a quadratic rate, nearly doubling the number of correct decimal digits at each iteration. This works in a fashion similar to the secant method, except the x-intercept is computed from the function value and its derivative, not from points at the ends of an interval.
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We assume you want the solution to ...
x^2 -5 = √(x -2)
This is conveniently reformulated to ...
x^2 -5 -√(x -2) = 0
so that we're seeking a zero of the function on the left.
The attached shows several iterations of the secant method. The function g is used to compute the x-intercept point on the line between (x1, f(x1)) and (x2, f(x2)). Then f(g( )) computes the function value at that x-intercept. We have used that to decide what the next interval limits (x1, x2) could be.
Our final guess at x is ...
x ≈ 2.367796 . . . . secant method
which gives an error less than 10^-5 between the left- and right-sides of the equation.
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Additional comments
At the bottom of the attachment is the Newton's Method iterator. The nice thing about this graphing calculator is that it displays the result of computing h(x) even as the value of x is being typed. This means you can type what you see and achieve full calculator precision for the solution in basically one pass. This method gives x = 2.36779701571.
Some hand-held graphing calculators have a derivative function built-in, so it is practical to use them to perform Newton's method iteration in substantially the same way.