To find the critical point we need t 2 4t 1 0 using

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Unformatted text preview: the absolute maximum and absolute minimum of f on the given interval. 1 − t 2 + 4t (t 2 + 1) a. f ( x) = x 4 − 2 x 2 + 3 Solution: 2 Since the numerator is always greater than zero. To find the critical point, we need − (t 2 − 4t − 1) = 0 , using quadratic equation 1 on [ - 1, 2] Hung Nguyen Problem Set 10 Solution 4.1 Minimum and Maximum 3 2 b. f ( x) = 2 x − 3x − 1 Solution: GE 207C on [- 2 0] f ( x) = 2 x 3 − 3x 2 − 1 f ' ( x) = 6 x 2 − 6 x = 6 x(x − 1) = 0 So the critical points are x = [0, 1] f (−2) = −29; f (0) = −1; f (1) = −2 So...
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