Section_4.2-The Mean_Value_Theorem - Section 4.2 The Mean Value Theorem 2010 Kiryl Tsishchanka The Mean Value Theorem THEOREM(The Extreme Value

# Section_4.2-The Mean_Value_Theorem - Section 4.2 The...

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Section 4.2 The Mean Value Theorem 2010 Kiryl Tsishchanka The Mean Value Theorem THEOREM (The Extreme Value Theorem): If f is continuous on a closed interval [ a,b ] , then f attains an absolute maximum value f ( c ) and an absolute minimum value f ( d ) at some numbers c and d in [ a,b ] . THEOREM (Fermat’s Theorem): If f has a local maximum or minimum at c, and if f ( c ) exists, then f ( c ) = 0 . THEOREM (Rolle’s Theorem): Let f be a function that satisfies the following three hypotheses: 1. f is continuous on the closed interval [ a,b ] . 2. f is differentiable on the open interval ( a,b ) . 3. f ( a ) = f ( b ) Then there is a number c in ( a,b ) such that f ( c ) = 0 . Proof: There are three cases: Case I: f ( x ) = k, a constant. Then f ( x ) = 0 , so the number c can be taken to be any number in ( a,b ) . Case II: f ( x ) > f ( a ) for some x in ( a,b ) . By the Extreme Value Theorem, f has a maximum value somewhere in [ a,b ] . Since f ( a ) = f ( b ) , it must attain this maximum value at a number c in the open interval ( a,b ) . Then f has a local maximum at c and, by hypothesis 2, f is differentiable at c. Therefore, f ( c ) = 0 by Fermat’s Theorem. Case III: f ( x ) < f ( a ) for some x in ( a,b ) . By the Extreme Value Theorem, f has a minimum value somewhere in [ a,b ] . Since f ( a ) = f ( b ) , it must attain this minimum value at a number c in the open interval ( a,b ) . Then f has a local minimum at c and, by hypothesis 2, f is differentiable at c. Therefore, f ( c ) = 0 by Fermat’s Theorem. EXAMPLE: Prove that the equation x 5 + 7 x 2 = 0 has exactly one real root. 1
Section 4.2 The Mean Value Theorem 2010 Kiryl Tsishchanka EXAMPLE: Prove that the equation x 5 + 7 x 2 = 0 has exactly one real root. Solution: We first show that the equation x 5 + 7 x 2 = 0 has a root. To this end we will use the Intermediate Value Theorem. In fact, let f ( x ) = x 5 +7 x 2 . Then f (0) = 0 5 +7 · 0 2 = 2 and f (1) = 1 5 + 7 · 1 2 = 6 . So, f
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