Transcription of Unit 5: Intermediate value theorem
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Unit 5: Intermediate value theorem
INTRODUCTION TO CALCULUSMATH 1 AUnit 5: Intermediate value solutions tog(x) =h(x) is equivalent to find solutionsf(x) =g(x) h(x) = :Iff(a) = 0, thenais called arootoff. Forf(x) = sin(x)for example, there are roots atx= 0,x= . are a few examplesExample:Find the roots off(x) = 4x+ : we setf(x) = 0 and solve forx. In this case 4x+ 6 = 0 and sox= 3 :Find the roots off(x) =x2+ 2x+ : Becausef(x) = (x+ 1)2the function has a root atx= 1. In generalx2+bx+chas :Find the roots off(x) = (x 2)(x+ 6)(x+ 3).Answer:Since thepolynomial is factored already, it is easy to see the rootsx= 2,x= 6,x= :f(x) = 12 +x 13x2 x3+x4. Find the roots off. There is no try (12 = 3 4 is a hint). We seex= 1,x= 3,x= 4,x= 1 are the :The functionf(x) = exp(x) does not have any :The functionf(x) = log|x|= ln|x|has rootsx= 1 andx= :f(x) = 2x 16 has the rootx= value theorem of continuous on the interval [a,b]andf(a),f(b) have different signs, then there is a root offin (a,b).
5.4. The following is an application of the intermediate value theorem and also provides a constructive proof of the Bolzano extremal value theorem which we will see later. Fermat’s maximum theorem If fis continuous and has f(a) = f(b) = f(a+ h), then fhas either a local maximum or local minimum inside the open interval (a;b). 5.5.
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