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Unformatted text preview: ntinuous function (can
you explain why?) and
F (0) = e0 + 3(0)2 − 4 = −3 < 0
F (1) = e + 3 − 4 > 0.
Therefore, the IVT shows that there is at least
one point c ∈ [0, 1] such that F (c) = 0. That is
ec + 3c2 − 4 = 0. 1
If we wanted a better idea of the location of c, we could bisect the original interval [0, 1] at the midpoint 2 , and evaluate
F ( 2 ). If F ( 1 ) < 0, then since F (1) > 0, we would have
a solution in the new interval [ 1 , 1].
Otherwise, if F ( 1 ) > 0, c would be located in
the interval [0, 1 ].
We can then bisect the new interval, test the
midpoint and repeat the procedure. In general, we now have an algorithm called the Bisection Method for ﬁnding approximate solutions of equations. This
method is illustrated in the next section. ii Copyrighted by B.A. Forrest (firstname.lastname@example.org) 3 Newton’s Method in Maple Bisection Method Suppose that we want to ﬁnd an approximate solution to the equation
f (x) = g (x)
where both f (x) and g (x) are continuous functions of x.
Step 1: Form the new continuous function
F (x) = f (x) − g (x)
Step 2: Find two points a0 < b0 such that either F (a0 ) > 0 and F (b0 ) < 0, or F (a0 ) < 0 and F (b0 ) > 0.
The IVT now shows us that there is a c between a0 and b0 such that F (c) = 0.
Step 3: Find the midpoint
d= a0 + b0
2 and evaluate F (d).
Step 4: If F (a0 ) and F (d) have opposite signs (i.e., one negative and one positive), then let a1 = a0 and b1 = d.
Otherwise, let a1 = d and b1 = b0 to obtain a new interval [a1 , b1 ] which contains a solution to the equation. We also have
that b1 − a1 = 2 (b0 − a0 ).
Step 5: Repeat steps 3 and 4 to obtain new intervals [a2 , b2 ], [a3 , b3 ],. . . ,[an , bn ], each of which contains a solution to the
equation. Then there is a c such that F (c) = 0 in each interval [ak , bk ].
Example [Bisection Method using Maple]
Use the Bisection Method to approximate the root of the polynomial f (x) = x5 + x − 1.
Start Maple. Begin a new worksheet by entering the restart: command. Enter the function f (x) in Maple.
[> f := x -> x∧5 + x - 1;
To get an idea of whether this function has a solution (root), look at plots of f (x) in Maple. Notice that we begin by using
a large set...
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