Unformatted text preview: c ∈ (1 , 5) so that equation (1) holds. To ﬁnd c we note that f (5)f (1) 51 = 1204 = 30 , and that f ( x ) = 3 x 21 . So we need to ﬁnd c ∈ (1 , 5) for which f ( c ) = 30, or 3 c 21 = 30 = ⇒ 3 c 2 = 31 = ⇒ c = ± s 31 3 . Clearly c =s 13 3 is not in the interval (1 , 5). The other solution satisﬁes c = s 31 3 ≈ 3 . 21 ∈ (1 , 5) . So the answer is c = s 13 3 ....
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 Fall '10
 KIHYUNHYUN
 Calculus, Topology, Intermediate Value Theorem, Mean Value Theorem, Metric space, Closed set

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