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Consider the following equation.

Approximate the solution to the equation above using three iterations of successive approximation. Use the graph below as a starting point.

Consider the following equation. Approximate the solution to the equation above using-example-1
User Radost
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1 Answer

3 votes

Answer:

A. x ≈ 25/16

Explanation:

You want the solution to 5^(-x) +7 = 2x +4 using 3 iterations of successive approximation.

Bisection method

There are many methods of successive approximation. We presume from the answer choices that you want to use the bisection method. Here, it can start with the interval [1, 2], bounded by the integers either side of the x-coordinate of the point where the curves intersect.

The bisection method has you evaluate the function at the midpoint of the interval. The midpoint of the interval will replace the endpoint whose function value has the same sign, thus reducing the interval width by half.

Intervals

The attached table shows the intervals containing the solution to the equation are reduced after 3 iterations to ...

[1, 2] ⇒ [3/2, 2] ⇒ [3/2, 7/4] ⇒ [3/2, 13/8]

The midpoint of this last interval is (12/8 +13/8)/2 = 25/16. This is our approximation of the solution.

x ≈ 25/16

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Additional comment

We find it convenient to rewrite the equation to the form f(x) = 0. Then we are looking for the x-intercept of the function f(x). For the function we have chosen, the function value is positive at the left end of the interval, and negative at the right end. Thus, we can choose which interval bound to replace based on the sign of the function value. This simplifies the mental effort required.

If you do the arithmetic using separate functions for the left- and right-sides of the equation, then you need to compare those values and choose the interval end to change based on that comparison. We find that to be extra work.

The bisection method, on average, adds about 1/3 of a decimal place to the precision of the solution with each iteration. Other iteration methods can converge much more quickly. For example, Newton's method iteration can get a 6 decimal place answer in two iterations starting from x ≈ 1.5.

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Consider the following equation. Approximate the solution to the equation above using-example-1
User Onlyjus
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