M408C: Derivatives and Graphing & HorizontalAsymptotesOctober 7, 2008Important concepts: First Derivative Test, concave up/concave down, concavity test, inflectionpoints, Second Derivative Test, determining long-term behavior of functions1.2.3.14. (4.3.12) For f(x) = x2/(x2+ 3), find the intervals on which f is increasing or decreasing. Findalso the local max and min values of f and find the intervals of concavity and the inflection points.Solution: A quick computation using the quotient rule tells us f0(x) = 6x/(x2+ 3)2. Thusf(x) = 0 only at x = 0. Now f0(x) > 0 for x > 0 and f0(x) < 0 for x < 0. Thus f is increasing forx > 0 and decreasing for x < 0. We next use the Second Derivative Test to classify local max andmin:f00(x) =6 · (x2+ 3)2− 6x ·2(x2+ 3)(2x)(x2+ 3)4=6(x2+ 3)(x2+ 3 −x · 2 · 2x)(x2+ 3)4=6(x2+ 3 − 4x2)(x2+ 3)3=6(3 − 3x2)(x2+ 3)3=18(1 − x2)(x2+ 3)3=18(1 − x)(1 + x)(x2+ 3)3.Now f00(0) = 18/27 = 2/3 > 0, so f has a local minimum of f(0) = 0 at x = 0. To determine theintervals of concavity, we need to determine the zero e s of f00and see how f00changes to the left andright of these zeroes. Clearly f00(x) = 0 for x = ±1. Observe that f00(x) < 0 for x > 1, f00(x) > 0for −1 < x < 1, and f00(x) < 0 for x < −1. Thus f is c oncave downward on (−∞, −1) and (1, ∞)and concave upward on (−1, 1). The inflection points are just (±1, 1/4) since f(1) = 1/4 = f(−1).5. (4.3.26) Suppose f (3) = 2, f0(3) =12, and f0(x) > 0 and f00(x) < 0 for all x. Sketch a possiblegraph for f . How many solutions does the equation f(x) = 0 have? Why? Is it possible thatf0(2) =13? Why?Solution: f0(x) > 0 for all x means that f is always increasing. f00(x) < 0 for all x means thatf is always concave downward. We know f(3) = 2 and f0(3) =12. So f could look like a smoothcurve that goes through (3, 2) and sits below the tangent line y −2 =12(x −3) at (3, 2). I will leaveit to you to draw the graph from this description.Now, the tangent line at (3, 2) can be written y =12x + 2. The x-intercept is thus at x = −1.f is concave down always, so we know that f has to be below the x-axis at x = −1. Thus f mustcross the x-axis at least once (between −1 and 3). But since f0(x) > 0 for all x, we know f is alwaysincreasing, so f can only cross the x-axis at most once. Thus it must cross the x-axis exactly once,in which case there is only one solution to f(x) = 0.Observe that f00(x) < 0 for all x means that f0is decreasing (i.e. as x gets larger, f0(x) getssmaller). In other words, x1< x2implies f0(x1) > f0(x2). Now 2 < 3, so f0(2) > f0(3). But1/3 < 1/2, so f0(2) 6=13.6. (4.4.30) Find limx→∞√x sin1x.Solution: Let t = 1/x. Thenlimx→∞√x sin1x= limt→0+1√tsin t = limt→0+t√tsin tt= limt→0+√tsin tt= 0 · 1 =
View Full Document
Unlocking...