4.2.1-eg-1 - c ∈ (1 , 5) so that equation (1) holds. To...

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Example Consider the function f ( x ) = x 3 - x, x [1 , 5] . Verify that the Mean Value Theorem holds in this case, and show that there is a c (1 , 5) for which f 0 ( c ) = f (5) - f (1) 5 - 1 . (1) Solution: The function f ( x ) = x 3 - x is differentiable on ( -∞ , ), so it is continuous on the closed interval [1 , 5] and differentiable on the open interval (1 , 5), the two conditions required for the results of the Mean Value Theorem to hold. From the Mean Value Theorem we are guaranteed the existence of at least one
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Unformatted text preview: c ∈ (1 , 5) so that equation (1) holds. To find c we note that f (5)-f (1) 5-1 = 120-4 = 30 , and that f ( x ) = 3 x 2-1 . So we need to find c ∈ (1 , 5) for which f ( c ) = 30, or 3 c 2-1 = 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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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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