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Question:

chowzor
chowzor
Student
College - Freshman

Show that function F(x)=(x-a)^2(x-b)^2+x takes on the value (a+b)/2 for some value of x

This is striaght of of my 1st year calculas textbook, a few of us are having issues the prof tried to help us out, but he is hard to understand, and doesn't know english well,

I know that;

F(?)=(a+b)/2

F(a)=a

F(b)=b

Our prof said to define g(x)=f(x)-(a+b)/2; and g(x)=0

then to apply IVT, im really lost

 

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Posted by chowzor on Thursday October 15, 2009 at 3:17 PM and tagged with math.

Answers:

  1. kjcdb8er
    kjcdb8er Teacher

    eNotes Editor

    The Intermediate Value Theorem (abbreviated as IVT) says that if a continuous function takes on two values A and B at points a and b, it also takes on every value between A and B at some point between a and b.

    Assume that a < b, without loss of generality. Note that a < (a+b)/2 < b (the important point is that (a+b)/2 lies between a and b)

    Note that F(a) = a < (a+b)/2 < b = F(a).

    The function F(x) takes on every value between a and b, by the IVT. (a+b)/2 is a value between a and b. Thus, F(c) = (a+b)/2 for some number c.

    I don't seen the advantage of defining a new function g(x). The method of applying the IVT is the same.

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    Posted by kjcdb8er on Thursday October 15, 2009 at 3:41 PM