select the correct location on the table. Consider the following equations.f(x)=x^3+3x^2-x-6. g(x)= x^2-3x-2. Approximate the solution to the equation f(x)=g(x) using 3 iterations of successive approximation. Use the graph as a starting point. NO LINKS!! ​

select the correct location on the table Consider the following equationsfxx33x2x6 gx x23x2 Approximate the solution to the equation fxgx using 3 iterations of class=

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Answer:

  15/16

Step-by-step explanation:

For the purpose here, we define a starting value as the midpoint of the interval between the integers on either side of the solution. Those integers are 0 and 1, so our starting value for an approximation of x is 1/2.

Each iteration will evaluate the difference f(x)-g(x) at this approximation. If the difference has the sign that matches the difference at the left end of the interval, then the approximation replaces the left end. If the sign matches the difference at the right end, then the approximation replaces the right end. The iteration ends with an approximation that is the midpoint of the new interval.

The attachment shows that our intervals start at [0, 1] and proceed through [1/2, 1], [3/4, 1] to [7/8, 1]. The approximation at the end of the 3rd iteration is x ≈ 15/16.
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Additional comment

The way this iterative process is defined, the denominator of the n-th approximation will be 2^(n+1). That is, the third iteration will result in a fraction with a denominator of 2^(3+1) = 2^4 = 16, as in 15/16.

As always with repetitive calculations, a graphing calculator or spreadsheet can reduce the work.

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