Using the Intermediate Value Theorem to Approximation a Solution ...
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Using the Intermediate Value Theorem to Approximation a Solution ...
Using the Intermediate Value Theorem to Approximation a. Solution to an
Equation. “Approximate a solution to the equation e−x2−1 = sin(x) to within 0.2.
Using the Intermediate Value Theorem to Approximation a Solution to an Equation
“Approximate a solution to the equation e−x
2
−1
= sin(x) to within 0.2.” We solve this as follows:
1. We first move all the terms to one side of the equation, so that we get an equation of the form “f (x) = 0”: 2 e−x −1 − sin(x) = 0 | {z } f (x)
Why do we do this? Since the IVT can tell us if a continuous function equals a certain value on a closed interval, we need to have (continuous function)=(some number). We know our f (x) is continuous by the theorems on pages 116-119 in book, and we’ll find closed intervals in the next steps. Note: It doesn’t matter which constant we have on the right, but 0 is typically the most convenient value since we can simply look at whether f (x) is positive or negative when trying to narrow down the interval from the next step. So, even if there is a constant on the right, we’ll go ahead and move it to the other side and include it in f (x). 2. We get a closed interval by plugging values of x into f until we get one positive and one negative: f (−1) = +0.9768 · · · = (+),
f (1) = −0.7061 . . . = (−)
If we hadn’t gotten a positive and a negative from the first two x-values, we would just keep trying until we had one of each. By the IVT, we know f has at least one zero somewhere in the interval [−1, 1]. 3. Now we narrow down the interval by evaluating f at values from the interval and seeing whether we get a positive or negative value. For example, f (0) = +0.3678... so we know there’s a zero in the interval [0, 1]. We repeat this process until we narrow down the interval to a width of 0.2 (the desired accuracy): f (0.5) = −0.192 . . . = (−), so we narrow to [0, .5]. Then f (0.3) = +0.040 . . . = (+), so we narrow to [.3, .5]. Now this interval has width 0.2, so our approximate solution is x = 0.4 (we could have picked any value in the interval - the midpoint is as good as any). Typically we’ll want a more accurate approximation, so more “narrowing” steps will be required in step 3. Also, remember to put your calculator in radian mode if your equation involves trig functions.
