Continuity, IVT and the Bisection Method

Math 131 – Continuity, the Intermediate Value Theorem, and the Bisection Method(09/14/2011)The graph of a continuous function is connected; that is, there are no holes or breaks in thegraph; the graph is in one piece. Another way of putting this is to say its behavior at a point c issimilar to its behavior near the point c. We characterize a function’s behavior near a point c bytalking about the limit of the function as x approaches c; hence the following definition.Definition: A function f ( x)is continuous at a point c if and only ifa. f ( c ) is definedb. lim ( )f x exists andx → clim f x = f c (the two are equal)x → cc. ( ) ( )Observe that this definition says three things: the function exists at c, the limit exists at c, and thetwo are the same. We extend this definition to continuity on an interval.Definition: A function f ( x)is continuous on an open interval (a,b) if and only if f ( x)iscontinuous at all points on the open interval (a,b).Definition: A function f ( x)is continuous on the closed interval [a,b] if and only if f ( x)iscontinuous on the open interval (a,b) and the left and right hand limits lim f ( x) f ( a)lim+x→b( ) ( )f x = f b exist and equal f at the endpoints.−x→a= andContinuous functions are well behaved and have many useful properties. One property is that acontinuous function on a closed interval [a, b] takes on all intermediate values betweenf ( a)and f ( b ) . Thus the following theorem:Intermediate Value Theorem (IVT): If f ( x)is continuous on the closed interval [a, b] and if Lis a value between f ( a)and f ( b ) then there is a point c between a and b such that f ( c)= L .f(b)(b, f(b))Lf(a)ac(a, f(a))b

Exercises: Using your calculator fill in a table similar to that above that uses the BisectionMethod to find the zeros for the following functions.21. Find the zero for the function f ( x) = x − 5 which is 5 . Note that ( )f ( 3)> 0 . Use 5 iterationsf 2 < 0 andA f(A) B f(B) MidPt=(A+B)/2 f(MidPt)22. Find the zero for the function f ( x) = x − 10 which is 10 ) . Note that ( )and f ( 3.5)= 2.25 > 0 . Use 5 iterationsf 3 = − 1

34. Find the zero for the function f ( x) = x − 10 which is 3 10 . Note that ( )f ( 2.2)= 0.648 > 0 . Use five iterations.f 2 = − 2 < 0 andA f(A) B f(B) MidPt=(A+B)/2 f(MidPt)5. Find the zero for the function ( )2so-called “golden ratio”. Note that f ( 1)= −1and f ( )f x = x − x − 1 which is the value of2 = 1. Use 5 iterations1+5Φ =2A f(A) B f(B) MidPt=(A+B)/2 f(MidPt)therevised 09/14/2011