Answer to Question #331328 in Calculus for Ramon

Question #331328

Use the bissection method to approximate the root of f(x)=2x²-1 in the interval . Let ε=0.1 be the margin of error of approximation.

1
Expert's answer
2022-04-21T04:12:01-0400

At first, we need to find the interval, where the roots are situated. "f(x)" is a quadratic function. "f(0)=-1" and "f(1)=f(-1)=1". Since the function is continuous, it intersects "x" -axis twice on the interval "[-1,1]". Consider the interval "[0,1]". The function intersects this interval once. I.e., the equation "f(x)=0" has one root. We denote this root by "x_1". Then, we set "x_2=-x_1". It will be the second root. It follows from the fact that "f(x)=f(-x)."

We start the bisection method:

  1. Divide the interval "[0,1]" into two parts: "[0,0.5]" and "[0.5,1]". Consider "f(0.5)=-0.5." From the value of "f(0.5)" we conclude that the root is in the interval "[0.5,1]".
  2. Divide the interval "[0.5,1]" into two parts: "[0.5,0.75]" and "[0.75,1]". Consider "f(0.75)=0.125". Thus, the root is in the interval "[0.5,0.75]".
  3. Divide the interval "[0.5,0.75]" into two parts: "[0.5,0.625]" and "[0.625,0.75]". Consider "f(0.625)=-0.21875". Thus, the root is in the interval "[0.625,0.75]".
  4. Divide the interval "[0.625,0.75]" into two parts: "[0.625,0.6875]" and "[0.6875,075]". Consider "f(0.6875)=-0.0546875". Thus, the root is in the interval "[0.6875,075]". We can stop the method, because the length of the intervals is less than "0.1". As an approximate value we set: "x_1=\\frac{0.6875+0.75}{2}\\approx0.7188"

Thus, the approximate value for one of roots is: "x_1\\approx0.7188". The approximate value for the second root is: "x_2\\approx-0.7188". We point out that exact values of roots are: "\\frac{1}{\\sqrt{2}}\\approx0.7071" and "-\\frac{1}{\\sqrt{2}}\\approx-0.7071".


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