MATH 1009 Midterm: Test 4 2016 Fall Solutions
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[ 5 marks] find the absolute maximum value and the absolute minimum value of the. The domain of f is the set of all real numbers. f(cid:48)(x) = ((2x2 + 3) 1)(cid:48) = (2x2 + 3) 2(2x2 + 3)(cid:48) = (2x2 + 3)2 = 0 when x = 0. Since the derivative is de ned for all real x (the denominator is never zero), the function has one critical number x = 0, which belongs to the interval [ 3, 1]. It remains to compare the value of f at the endpoints of the interval [ 3, 1] and at the critical number: f ( 3) = and the absolute minimum value is f ( 3) = [4 marks] find all the in ection points of the function f (x) = x3 3x2 + 12x + 10. (give both the x- and the y-coordinates. ) Thus, the absolute maximum value is f (0) =
