Math 31A 2009 11 03 MATH 31A DISCUSSION JED YANG 1 More Applications of the Derivative 1 1 Recall 1 1 1 Concavity If f x is increasing or f x 0 then f is concave up at x If f x is decreasing or f x 0 then f is concave down at x 1 1 2 Inflection If f c 0 and f x changes sign at x c then f x has a point of inflection at x c 1 1 3 Second Derivative Test Let f be differentiable and c a critical point f c 0 a If f c 0 then f c is a local minimum b If f c 0 then f c is a local maximum c If f c 0 then it is inconclusive f c may be a local min max or neither 1 2 Exercise 4 4 24 Find the critical points of f x sin2 x cos x x in 0 and use the Second Derivative Test to determine whether each corresponds to a local minimum or maximum Solution Notice f x 2 sin x cos x sin x Setting f x 0 and solving we get sin x 2 cos x 1 0 so x 0 3 Now f x 2 cos2 x 2 sin2 x cos x So f 0 1 f 3 32 f 3 yielding local minima at x 0 and maximum at x 3 1 3 Exercise 4 4 53 If f c 0 and f c is neither a local min or max must x c be a point of inflection This is true of most reasonable examples but it is not true in general Let 2 x sin x1 for x 6 0 f x 0 for x 0 Use the limit definition of the derivative to show that f 0 exists and f 0 0 Show that f 0 is neither a local min nor max Show that f x changes sign infinitely often near x 0 and conclude that f x does not have a point of inflection at x 0 Solution Recall Exercise 3 7 92 from 10 20 x 0 Recall f x limh 0 f x h f So by definition f 0 limh 0 f h f h h 1 1 limh 0 h sin h Using Squeeze Theorem and h h sin h h we get that f 0 0 Away from x 0 we can use the formula and get f x 2x sin x1 x2 cos x1 1 x12 2x sin x1 cos x1 Now limx 0 f x does not exist since limx 0 2x sin x1 0 by Squeeze Theorem but limx 0 cos x1 does not exist Math 31A Yang 2 1 4 Exercise 4 5 59 1 5 Exercise 4 5 68 1 6 Bonus Question Assume f x exists and f x 0 for all x Show that f x cannot be always negative Solution If f x 0 then f x 0 a contradiction So there exists b such that f b 6 0 Consider the tangent line at x b to f x It is given by the equation y f b x b f b Consider g x f x f b x b f b Notice that g x f x f b and g x f x So g b g b 0 and g x 0 for all x Hence g x is increasing In particular g x 0 for x b and g x 0 for x b If x b then by MVT we get g x g b g c x b for some c in the interval b x In otherwords since c b we have that g c 0 nad x b 0 hence g x g b 0 Similarly if x b we get g c 0 x b 0 so g x g b 0 as well We thus conclude g x g b for all x So f x f b x b f b for all x Since f b 6 0 there exists x far enough from the origin such that f b x b f b 0 as desired