f(x)=x^3+3
g(c)=x^2+2
Approximate the solution to the equation f(x)=g(x) using three iterations of successive approximation. Use this graph as a starting point. (It’s not x= -7/8)

In the graphing tool, choose the custom option in the Relationship menu to graph the functions f(x) = x^3 + 3 and g(x) = x^2 + 2.

Adjust the zoom level of the graph so you can see the point where the two graphed functions intersect. Then, left-click on the point where the functions intersect. The values of the point you click on, rounded to the nearest hundredth, will appear for about 2 seconds.

Note: If you’re not using a mouse (or a mouse with left-click ability), perform the equivalent zoom-in action on your device to see the intersection point values rounded to the nearest hundredth.

Then, approximate (to the nearest hundredth) the solution of f(x) = g(x) from part A of this question.

15 points PLZZZZZ URGENTTT! Urgent maximum wait is 24 hours!

It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...

evaluating the function at the midpoint of the bracketing interval
choosing a smaller bracketing interval
identifying the x-value known to be closest to the solution

Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.

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(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.

For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.

For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.

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(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.

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(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.

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See the attached table for function values.

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Comment on bisection iteration

Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.

Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.