Chapter6

# Chapter6 - Chapter 6 The Mean Value Theorem 6.0 The Theorem...

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Unformatted text preview: Chapter 6 The Mean Value Theorem 6.0 The Theorem Problem . If a car completes a 110km trip in one hour along a road with speed limit of 100 km h , show that the car exceeded the speed limit at some point in that hour of travel. (The car is allowed to travel backwards.) This problem may sound trivial; however, is it really the case that knowledge of the average velocity translates to knowledge of the instantaneous velocity? To solve the problem, we let s ( t ), 0 t 1 represent the signed distance travelled (in kilometers) t hours after the start of the trip ( t = 0), and we let v ( t ) = s ( t ) be the instantaneous velocity at time t . The speed of the vehicle is given by | v | ( t ) := | v ( t ) | = | s ( t ) | . The average velocity of the one-hour trip is given by s (1)- s (0) 1- = s (1) = 110 km h . We must find some t (0 , 1) so that v ( t ) = s ( t ) = 110 km h . Theorem 6.0.1 [Rolles Theorem] . Suppose that f : S R , where S R , and that [ a,b ] S with a &lt; b . Also suppose that f ( x ) is continuous on [ a,b ] and differentiable on ( a,b ) . Assume that f ( a ) = 0 = f ( b ) ; then there exists c ( a,b ) such that f ( c ) = 0 . Proof. By the extreme value theorem, f ( x ) attains both its global maximum and minimum on [ a,b ]. If they both occur at the endpoints, then f ( a ) = 0 = f ( b ) satisfies 0 f ( x ) 0 for all x [ a,b ] i.e. , f ( x ) = 0 is a constant function on all of [ a,b ]. Hence any c ( a,b ) will satisfy f ( c ) = 0. If a global extremum for f ( x ) on [ a,b ] occurs at c ( a,b ), then c is a local extremum for f ( x ) and hence f ( c ) = 0 by theorem 5.4.2, as required. We now come to one of the most important theorems in calculus. 93 Theorem 6.0.2 [Mean Value Theorem] . Suppose that f : S R , where S R , and that [ a,b ] S with a &lt; b . Also suppose that f ( x ) is continuous on [ a,b ] and differentiable on ( a,b ) . Then there exists a c ( a,b ) such that f ( c ) = f ( b )- f ( a ) b- a . Proof. Let h : S R be defined by h ( x ) = f ( x )- h f ( a ) + f ( b )- f ( a ) b- a ( x- a ) i . Then h ( x ) is continuous on [ a,b ] by theorem 4.2.13 and continuous on ( a,b ) by theorem 5.1.1. Note that h ( a ) := f ( a )- f ( a ) + f ( b )- f ( a ) b- a ( a- a ) = f ( a )- f ( a ) = 0 and h ( b ) := f ( b )- f ( a ) + f ( b )- f ( a ) b- a ( b- a ) = f ( b )- [ f ( a ) + f ( b )- f ( a )] = 0 . By Rolles theorem, there exists some c ( a,b ) with h ( c ) = 0. But 0 = h ( c ) = f ( c )- f ( b )- f ( a ) b- a by theorem 4.2.13. Hence f ( c ) = f ( b )- f ( a ) b- a , as required. Now we are able to show something that we assumed to be obvious. Corollary 6.0.3. Suppose that f : S R , where S R , and that [ a,b ] is a nondegenerate interval such that [ a,b ] S . Also suppose that f ( x ) is continuous on [ a,b ] and differentiable on ( a,b ) . If f ( c ) = 0 for all x ( a,b ) , then there exists M R such that f ( x ) = M...
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## Chapter6 - Chapter 6 The Mean Value Theorem 6.0 The Theorem...

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