Example: Brazilian Capybaras Click here to open an associated Mathcad worksheet:

Problem. See previous example.

Solution (revisited). Notice that the algebraic problem of solving

2 + (0.002) t 3 = (1.1) t

(abandoned previously) is the same as solving

2 + (0.002) t 3 – (1.1) t = 0 .

That is, it is the same as finding a root of the function  f(t) = 2 + (0.002) t 3 – (1.1) t .

Using our graphical observation that the populations appear to be equal somewhere between  t = 60  and  t = 70, we might use the formula for  f  to calculate approximate values of  f(60)  and  f(70) :

Since the value of the function changes sign between  t = 60  and  t = 70, we've confirmed that a root lies somewhere in-between. Suppose we look half way in-between:

Now we see that the value of  f  changes sign between  t = 65  and  t = 70, and we can conclude that the root must lie somewhere in this interval. Computing the value half way across this new interval leads to another table entry:

We now see that the root is in the interval between  t = 67.5  and  t = 70.

The method is clear enough: Keep subdividing the one interval on which  f  changes sign and compute a new value for the table at the midpoint. Eventually, after many iterations of this simple step, we will be able to compute the root to any degree of accuracy. This may seem tedious, but its algorithmic nature makes it an ideal method for use on calculators or computers, which can perform tedious calculations very quickly.

After five more iterations, the root is bracketed between  t = 67.27  and  t = 67.34. (Try it!) We may conclude that, to one decimal place, the root is at  t = 67.3. This much accuracy could have been achieved fairly quickly by graphing and zooming. Unlike graphical methods, however, there is no limit to the degree of accuracy that we may obtain by using the numerical method over and over again.

 
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