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Unformatted text preview: CM221A ANALYSIS I Solutions to Exercise Sheet 7 1. Find the maximum and minimum values of the function f ( x ) = 2 x 5- x 3 on the interval [0 , 1], and sketch the graph of the function on this interval. Solving f ( x ) = 0 , we find the only solution in the interval, at q 3 / 10 . Now we look at the values of the function at the two end points as well, to find out if the maximum and minimum occur there. Since f (0) = 0 and f (1) = 1 , the global maximum occurs for x = 1 and the global minimum for x = q 3 / 10 . 2. Prove that the derivative of the function f ( x ) = sin 2 ( x ) + x 2 only vanishes at one point in [- π,π ]. Find the maximum and minimum values of the function f ( x ) on the interval and sketch its graph. Since f (- x ) = f ( x ) we only need to consider x ≥ . The function f is positive on [- π,π ] except at x = 0 where f (0) = 0 . f ( x ) = 2 x + sin(2 x ) implies f (0) = 0 . If < x ≤ π/ 2 then both terms in f ( x ) are positive. If π/ 2 ≤ x ≤ 2 π then f (...
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- Spring '09
- Calculus, Mathematical analysis, mean value