Math 409-502Harold P. [email protected] derivativeDefinition. A function f is differentiable at a point a if the limit limx→af (x) − f (a)x − aexists. The limit,if it exists, is called the derivative and is denoted by f0(a).Example 1. If f (x) = x2, then f0(5) = limx→5x2− 52x − 5= limx→5(x + 5) = 10.Example 2. If f (x) = |x|, then f0(0) does not exist. Indeed, limx→0+f (x) − f (0)x − 0= limx→0+1 = 1,but limx→0−f (x) − f (0)x − 0= limx→0−−1 = −1. So there is a right-hand derivative f0(0+) and there is aleft-hand derivative f0(0−), but they are not equal.Math 409-502 November 8, 2004 — slide #2The mean-value theoremTheorem. If f is a continuous function on a compact interval [a, b], and if f is differentiable at allinterior points of the interval, then there exists an interior point c for which f0(c) =f (b) − f (a)b − a.Example application. If the derivative of a function is identically equal to zero on an interval,then the function is constant on the interval.Proof. Fix a point a in the interval. Let b be any other point.By the mean-value theorem, there is a point c for which f (b) − f (a) = f0(c)(b − a) = 0. Sof (b) = f (a) for every point b.Math 409-502 November 8, 2004 — slide #3Proof of the mean-value theoremLet g be the difference between f and the line joining the points (a, f (a)) and (b, f (b)): g(x) =f (x) −µf (a) +f (b) − f (a)b − a(x − a)¶.We seek a point c for which g0(c) = 0.The function g is continuous on the compact interval [a, b], so g attains a maximum value anda minimum value. One of these must be attained at an interior point c because g(a) = 0 andg(b) = 0. We may suppose the maximum is attained at c.Then g(x) − g(c) ≤ 0 for all x, so when x − c > 0 we haveg(x) − g(c)x − c≤ 0, whence g0(c+) ≤ 0.If x − c < 0, theng(x) − g(c)x − c≥ 0, so g0(c−) ≥ 0.By hypothesis, the derivative g0(c) exists, so the one-sided derivatives are equal. Thus g0(c) = 0as required.Math 409-502 November 8, 2004 — slide #4Homework1. Read Chapter 14 (pages 196–204) and section 15.1 (pages 210–211).2. Do Exercise 14.1/3 on page 205.3. Do Exercise 15.1/4 on page 218.Math 409-502 November 8, 2004 — slide