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Announcements Topics: - finish section 5.4 (derivatives of inverse trig functions), work on sections 5.6 (second derivative), 5.7 (taylor polynomials), 6.1 (extreme values) and 6.4 (l’Hopital’s rule) * Read these sections and study solved examples in your textbook! Work On: - Practice problems from the textbook and assignments from the coursepack as assigned on the course web page (under the link “SCHEDULE + HOMEWORK”) Derivatives of Inverse Trig Functions d 1 (arcsin x) = dx 1- x 2 d 1 (arctan x) = 2 dx 1+ x Example 1: Differentiate. (a) f (x) = arctan(5x) + 5arctan x (b) g(x) = ln(arcsin(5x 3 +1)) Example 2: Prove d 1 (arctan x) = . 2 dx 1+ x The Second Derivative The derivative of the derivative is called the second derivative. d2 f the second derivative of f = f "(x) = 2 dx The Second Derivative f” provides information about f’ and f: When f’’ is positive, f’ is increasing, i.e., the rate at which f is changing is increasing. When f’’ is positive, the slopes of the tangents to the graph of f are increasing and the graph of f is concave up. The Second Derivative f” provides information about f’ and f: When f’’ is negative, f’ is decreasing, i.e., the rate at which f is changing is decreasing. When f’’ is negative, the slopes of the tangents to the graph of f are decreasing and the graph of f is concave down. The Second Derivative When the graph of f changes concavity at a point in the domain of f, this point is called an inflection point. Note: At an inflection point, f”=0 or f’’ D.N.E. The Second Derivative Example: Gamma Distribution g(x) = xe-x , x ³0 Find the first and second derivatives of g(x) and use them to sketch a graph. Approximating Functions with Polynomials Polynomials have many nice properties and are generally very easy to work with. For this reason, it is often useful to approximate more complicated functions with polynomials in order to simplify calculations. Linear functions are the simplest polynomials and can be used to represent a more complicated function in many situations. Linear Approximations The secant line connecting points (a, f(a)) and (b, f(b)) on the graph of f(x) provides a decent linear approximation to f(x) for x-values in the interval [a, b]. The tangent line to the graph of f(x) at (a, f(a)) provides the best linear (straight line) approximation to f(x) near x=a. Linear (or tangent line) approximation to f(x) around x=a: L(x) = f (a) + f '(a)(x - a) Linear Approximations Example: Let f (x) = x . Find the secant line approximation to f(x) for x-values between 1 and 4. Use this to approximate both 2 and 3 . Compare to the actual values. Using Technology: 2 »1.41421356237 3 »1.73205080757 Linear Approximations Example: Let f (x) = x . Find the tangent line approximation to f(x) at x=1. Use this to approximate both 2 and 3 . Compare to the actual values. Using Technology: 2 »1.41421356237 3 »1.73205080757 Quadratic Approximations We can obtain a more accurate approximation by using polynomials of higher degrees. The tangent line (linear) approximation matches the value and the slope of the function at x=1. The quadratic approximation matches the value, the slope, and the curvature (concavity) of the function at x=1. L(x) =1+ 12 (x -1) f (x) = x T2 (x) =1+ 12 (x -1) - 18 (x -1)2 Quadratic Approximations To find the quadratic approximation to a function f (x) at the base point a, we match the value, the first derivative, and the second derivative at the point a. Quadratic approximation to f (x) around x=a: f ''(a) T2 (x) = f (a) + f '(a)(x - a) + (x - a) 2 2 because the degree is 2 Quadratic Approximations Example: 1 . Let f (x) = 1+ x Find the quadratic approximation to f(x) around x=1. • f (x) = 1 x +1 The Taylor Polynomial Suppose the first n derivatives of the function f are defined at x=a. Then the Taylor polynomial of degree n matching the values of the first n derivatives is nth derivative The Taylor Polynomial Example: Find the 1st, 3rd, and 5th degree Taylor polynomials for f (x) = sin x near x=0. The Taylor Polynomial P1(x) = x P5 (x) = 5!1 x 5 - 3!1 x 3 + 1!1 x f (x) = sin x P3 (x) = - 3!1 x 3 + 1!1 x Maximum and Minimum Values f (c) is a global (absolute) maximum of f if f (c) ³ f (x) for all x in the domain of f . f (c) is a local (relative) maximum of f if f (c) ³ f (x) for all x in some interval around c. Maximum and Minimum Values f (c) is a global (absolute) minimum of f if f (c) £ f (x) for all x in the domain of f . f (c) is a local (relative) minimum of f if f (c) £ f (x) for all x in some interval around c. Extrema Identify the labeled points as local maxima/minima, global maxima/minima, or none of these. Extrema Identify the labeled points as local maxima/minima, global maxima/minima, or none of these. local max global max local max nothing special local min global min Extreme Values Notice: Extreme values occur at either a critical number of f or at an endpoint of the domain. (However, not all critical numbers and endpoints correspond to an extreme value.) Also note: By definition, relative extreme values do not occur at endpoints. Finding Local Maxima and Minima (First Derivative Test) Assume that f is continuous at c, where c is a critical number of f. If f’ changes from + to - at x=c, then f changes from increasing to decreasing at x=c and f(c) is a local maximum value. If f’ changes from - to + at x=c, then f changes from decreasing to increasing at x=c and f(c) is a local minimum value. If f’ does not change sign at x=c, then f doesn’t have an extreme value at x=c. Finding Local Maxima and Minima (First Derivative Test) Example: Use the first derivative test to find the local extrema of the following functions. (a) f (x) = x ln x x (b) g(x) = 2 1+ x Finding Local Maxima and Minima (Second Derivative Test) Assume that f’’ is continuous near c and f’(c)=0. If f’’(c)>0 then the graph of f is concave up at x=c and f(c) is a local minimum value. If f’’(c)<0 then the graph of f is concave down at x=c and f(c) is a local maximum value. If f’’(c)=0 or f”(c) D.N.E. then the second derivative test doesn’t apply and you have to use the other method. Application Assignment 43, #1 (modified): Consider the function f (t) = Ate - bt , where A, b > 0. (a) Find the critical number of f. Application Assignment 43, #1 (modified): (b) Use the second derivative test to determine if the critical number in part (a) corresponds to a local maximum, local minimum, or neither. Application Assignment 43, #1 (modified): (c) Determine the values of A and b such that f describes the graph given below. Extreme Value Theorem If f (x) is continuous for all x Î [a, b] , then there are points c1, c 2 Î [a, b] such that f (c1) is the global minimum and f (c 2 ) is the global maximum of f (x) on [a, b]. In words: If a function is continuous on a closed, finite interval, then it has a global maximum and a global minimum on that interval. Finding Absolute Extreme Values on a Closed Interval [a,b] 1. Find all critical numbers in the interval. 2. Make a table of values. The largest value of f(x) is the absolute maximum and the smallest value is the absolute minimum. Finding Absolute Extreme Values on a Closed Interval [a,b] Example: 1 2 3 Find the absolute extrema of g(x) = x (x - 2) on [-1, 1]. L’Hopital’s Rule Another application of derivatives is to help evaluate limits of the form where either or Idea: Instead of comparing the functions f(x) and g(x), compare their derivatives (rates) f’(x) and g’(x). L’Hopital’s Rule Suppose that f and g are differentiable functions such that is an indeterminate form of type 00 or ¥¥ . If g¢(x) ¹ 0 near a (could be 0 at a) then L’Hopital’s Rule Evaluate the following limits using L’Hopital’s Rule, if it applies. (a) (b) (c) (d) L’Hopital’s Rule Evaluate the following limits using L’Hopital’s Rule, if it applies. (a) (b)