T4_11_SecA_sols

T4_11_SecA_sols - MATH 1009 A TEST 4 SOLUTIONS 1 5 Marks...

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MATH 1009 - A TEST 4 - SOLUTIONS ———————————————————————————————————————- 1. [ 5 Marks] Find the absolute maximum value and the absolute minimum value of the function f ( x ) = 5 x 2 + 1 on the interval [ 1 , 2] . Solution: The domain of f is the set of all real numbers. f ( x ) = (5( x 2 + 1) 1 ) = 5( x 2 + 1) 2 ( x 2 + 1) = 10 x ( x 2 + 1) 2 = 0 when x = 0. Since the derivative is deﬁned for all real x (the denominator is never zero), the function has one critical number x = 0, which belongs to the interval [ 1 , 2]. It remains to compare the value of f at the endpoints of the interval [ 1 , 2] and at the critical number: f ( 1) = 5 / 2, f (2) = 1, f (0) = 5. Thus, the absolute maximum value is f (0) = 5 and the absolute minimum value is f (2) = 1. 2. [4 Marks] Find all the inﬂection points (if any) of the function f ( x ) = x 3 3 x 2 + x + 1. Solution:

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T4_11_SecA_sols - MATH 1009 A TEST 4 SOLUTIONS 1 5 Marks...

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