121616949-math.148

# 121616949-math.148 - both occur at an endpoint and since...

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134 Chapter 6 Applications of the Derivative much experimentation before you are convinced of the truth of this statement: Somewhere between t 0 and t 1 the slope is exactly zero, that is, somewhere between t 0 and t 1 the slope is equal to the slope of the line between the endpoints. This suggests that perhaps the same is true even if the endpoints are at different heights, and again a bit of experimentation will probably convince you that this is so. But we can do better than “experimentation”—we can prove that this is so. We start with the simplified version: THEOREM 6.25 Rolle’s Theorem Suppose that f ( x ) has a derivative on the interval ( a, b ), is continuous on the interval [ a, b ], and f ( a ) = f ( b ). Then at some value c ( a, b ), f ( c ) = 0. Proof. We know that f ( x ) has a maximum and minimum value on [ a, b ] (because it is continuous), and we also know that the maximum and minimum must occur at an endpoint, at a point at which the derivative is zero, or at a point where the derivative is undefined. Since the derivative is never undefined, that possibility is removed. If the maximum or minimum occurs at a point c , other than an endpoint, where f ( c
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Unformatted text preview: both occur at an endpoint, and since the endpoints have the same height, the maximum and minimum are the same. This means that f ( x ) = f ( a ) = f ( b ) at every x ∈ [ a, b ], so the function is a horizontal line, and it has derivative zero everywhere in ( a, b ). Then we may choose any c at all to get f ± ( c ) = 0. Perhaps remarkably, this special case is all we need to prove the more general one as well. THEOREM 6.26 Mean Value Theorem Suppose that f ( x ) has a derivative on the interval ( a, b ) and is continuous on the interval [ a, b ]. Then at some value c ∈ ( a, b ), f ± ( c ) = f ( b )-f ( a ) b-a . Proof. Let m = f ( b )-f ( a ) b-a , and consider a new function g ( x ) = f ( x )-m ( x-a )-f ( a ). We know that g ( x ) has a derivative everywhere, since g ± ( x ) = f ± ( x )-m . We can compute g ( a ) = f ( a )-m ( a-a )-f ( a ) = 0 and g ( b ) = f ( b )-m ( b-a )-f ( a ) = f ( b )-f ( b )-f ( a ) b-a ( b-a )-f ( a ) = f ( b )-( f ( b )-f ( a ))-f ( a ) = 0 ....
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