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Math 1000 Tutorial Quiz 7 Week 7 The solutions to Quiz 7 are below. I might be showing more steps than necessary, this is to aid your understanding. 1 1. Find the critical numbers of f (x) = x4 + 3x2 + 6. Solution: The derivative is given by f 0 (x) = 4x3 + 6x. First, find when f 0 (x) = 0. If 4x3 + 6x = x(4x2 + 6) = 0, then either x = 0 or 4x2 + 6 = 0. There is no real solution for the second, thus f 0 (x) = 0 only if x = 0. Second, the derivative is defined everywhere. Therefore, the critical number is x = 0. 2 2. Find the local and absolute maxima and minima of f (x) = x − x3 on [−1, 1]. Solution: (i) Find the critical numbers: q f 0 (x) = 1 − 3x2 , setting equal to zero we have x = ± 13 , both solutions are on the interval. f 0 (x) is defined everywhere on the interval, thus these two are the only critical numbers. (ii) Evaluate the function at the critical numbers and the endpoints of the interval: f (−1) = 0 f (1) = 0 q q q f ( 13 ) = 13 − ( 13 )3 > 0 q q q f (− 13 ) = − 13 − (− 13 )3 < 0 (iii) Comparing these values we get: q q q 1 There is a local and absolute maximum at ( 13 , 13 − 27 ). q q q 1 There is a local and absolute minimum at (− 13 , − 13 + 27 ). 1 3. (a) Does f (x) = x(x2 − x − 2) satisfy the criteria for the Mean Value Theorem on [−1, 1]? Explain why or why not. Solution: Because f (x) is a polynomial, it is continuous on [−1, 1] and differentiable on (−1, 1). Thus f (x) satisfies the criteria for the Mean Value Theorem on [−1, 1]. 1/ 2 (bonus) (b) Find a c ∈ (−1, 1) satisfying f 0 (c) = f (1)−f (−1) 1−(−1) for the above function. Solution: By part (a) we can apply the Mean Value Theorem, thus such a c exists. Now, f 0 (x) = 3x2 − 2x − 2, implying f 0 (c) = 3c2 − 2c − 2. Also, f (1)−f (−1) 1−(−1) = −2−0 2 = −1. Thus setting these equal, we have 3c2 − 2c − 2 = −1. Using the quadratic formula we get c = − 13 or c = 1. The second solutions is not in the interval, thus the c satisfying the conclusion of the Mean Value Theorem is c = − 13 in this case.