Math 409-502Harold P. [email protected] Club MeetingMonday, November 15 at 7:30 PMBlocker 156Speaker: Jeff Nash, Aggie former student, from Veritas DGCTitle: Math and Seismic ImagingFREE FOODMath 409-502 November 8, 2004 — slide #2The derivative and applicationsThe definition: f0(a) := limx→af (x) − f (a)x − aSome applications:• the mean-value theorem• l’Hˆopital’s rule• monotonicity; extreme valuesMath 409-502 November 8, 2004 — slide #3The mean-value theorem generalizedCauchy’s form of the mean-value theorem: Suppose f and g are continuous functions on [a, b]that are differentiable on (a, b). Then there exists a point c in (a, b) for whichµf (b) − f (a)b − a¶g0(c) =µg(b) − g(a)b − a¶f0(c),orf (b) − f (a)g(b) − g(a)=f0(c)g0(c)if the denominators are non-zero.Proof. Set h(x) =³f (b)− f (a)b−a´g(x) −³g(b)−g(a)b−a´f (x).A computation shows that h(a) = h(b), so by the ordinary mean-value theorem, there is apoint c for which h0(c) = 0. That reduces to the desired conclusion.Math 409-502 November 8, 2004 — slide #4A best-selling authorGuillaume Franc¸ois Antoine Marquis de L’Hˆopital(1661–1704)Author of the first calculus textbook:Analyse des infiniment petitsMath 409-502 November 8, 2004 — slide #5L’Hˆopital’s rule(s)If f and g are differentiable functions, and if limx→af (x)g(x)is a formally undefined expression of theform00or∞∞, then limx→af (x)g(x)= limx→af0(x)g0(x)if the second limit exists.Variations: one-sided limits; x → ∞.Example. limx→∞ln(x)xis formally∞∞. L’Hˆopital’s rule says that the limit equals limx→∞1/x1= 0.Example. limx→0+x ln(x) is formally 0 · (−∞). Rewrite as limx→0+ln(x)1/x. L’Hˆopital’s rule says that thelimit equals limx→0+1/x−1/x2= limx→0+(−x) = 0.Math 409-502 November 8, 2004 — slide #6Proof of one version of L’Hˆopital’s ruleTheorem. Suppose f and g are differentiable for large x and limx→∞f (x) = ∞ and limx→∞g(x) = ∞.If limx→∞f0(x)/g0(x) = ∞, then limx→∞f (x)/g(x) = ∞.Proof. Fix M > 0. We must find b such that f (x)/g(x) > M when x > b.By hypothesis, there is a such that f (a) > 0 and f0(c)/g0(c) > 2M when c ≥ a. Moreover, thereis b > a such that |g(a)/g(x)| < 1/2 when x > b.Apply Cauchy’s mean-value theorem when x > b to getf (x)− f (a)g(x)− g(a)=f0(c)g0(c)> 2M.Then f (x) > f (x) − f (a) > 2M(g(x) − g(a)), so f (x)/g(x) > 2M(1 −g(a)g(x)) > 2M(1 −12) = M.Math 409-502 November 8, 2004 — slide #7Homework1. Read sections 15.2–15.4, pages 212–217.2. Do Exercise 14.3/2 on page 206.3. Do Exercise 15.4/2 on page 219.Math 409-502 November 8, 2004 — slide