Chapter 4: More Applications of Differentiation

Chapter 4: More Applications of Differentiation Winter 2016 Department of Mathematics Hong Kong Baptist University 1 / 61

In the fall of 1972, President Nixon announced that, the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection. In this chapter we will examine several kinds of problems to which the differentiation can be applied: (1) evaluation of limits ( 4.3) (2) optimization problems ( 4.4) (3) graphing problems ( 4.5, 4.6) (4) approximation problems ( 4.9, 4.10) 2 / 61

4.3 Indeterminate Forms Recall that in 2.5, we showed that lim x 0 sin(x)/x = 1. We could not readily see this by substituting x = 0 into the function because both sin(x) and x are zero at x = 0. We call sin(x)/x an indeterminate form of type [0/0] at x = 0. The limit of such an indeterminate form can be any number, e.g., 0, k,, or does not exist. For instance, each of the quotients kx/x, x/x 3 and x 3 /x 2 is an indeterminate form of type [0/0] at x = 0, but kx lim x 0 x = k, lim x 0 x x 3 = and lim x 3 x 0 x 2 = 0. 3 / 61

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Indeterminate forms of type [0/0] are the most common. Many examples of the type [0/0] can be evaluated with simple algebra, typically by canceling common factors. For instance, lim x 1 x 2 1 x 2 3x + 2 = lim x + 1 x 1 x 2 = 2. In this section we develop another method called l Hôpital Rules for evaluating limits of indeterminate forms of the types [0/0] and [ / ]. The other types of indeterminate forms can usually be reduced to one of these two by algebraic manipulation and the taking of logarithms. 5 / 61

Theorem (The first l Hôpital Rule: for type [0/0]) Suppose that f and g are differentiable on the interval (a, b) and that g (x) 0 there. Suppose also that (i) (ii) Then lim f (x) = lim g(x) = 0, and x a+ x a+ lim x a+ f (x) g = L (where L is finite, or or ). (x) f (x) lim x a+ g(x) = L. Similar results hold if every occurrence of lim x a+ is replaced by lim x b or even lim x c where a < c < b. The cases a = and b = are also allowed. 6 / 61

Example 1: Evaluate (a) lim x 1 Solution: ln(x) x 2 ; (b) lim 1 x 0 2 sin(x) sin(2x) 2e x 2 2x x 2. 7 / 61

Example 1: Evaluate (a) lim x 1 Solution: (a) By the first l Hôpital Rule, lim x 1 ln(x) x 2 ; (b) lim 1 x 0 ln(x) x 2 1 = lim 1/x x 1 2x = lim x 1 2 sin(x) sin(2x) 2e x 2 2x x 2. 1 2x 2 = 1 2. (b) Repeatedly by the first l Hôpital Rule, we have lim x 0 2 sin(x) sin(2x) cos(x) cos(2x) 2e x 2 2x x 2 = lim x 0 e x 1 x sin(x) + 2 sin(2x) = lim x 0 e x 1 cos(x) + 4 cos(2x) = lim x 0 e x = 3. (still (still [ ] 0 ) 0 [ ] 0 ) 0 8 / 61

Example 2: Evaluate lim x 1+ x ln(x). Solution: x L Hôpital s Rule cannot be used to evaluate lim x 1+ ln(x) because this is not an indeterminate form. Note that the denominator approaches 0 and the numerator approaches 1 as x 1+. We have lim x 1+ x ln(x) =. Remark: Nevertheless, if we tried to use the l Hôpital Rule, it would lead to the WRONG answer lim x 1+ x ln(x)?! = lim x 1+ 1 1/x = 1. 9 / 61

( 1 Example 3: Evaluate lim x 0+ x 1 ). sin(x) Solution: Note that the indeterminate form here is of the type [ ]. Thus, l Hôpital s Rule cannot be applied directly. To solve the problem, we do the following transformation to turn it into the type [0/0]. Specifically, lim x 0+ ( 1 x 1 sin(x) ) sin(x) x = lim x 0+ x sin(x) = lim x 0+ cos(x) 1 sin(x) + x cos(x) sin(x) = lim x 0+ 2 cos(x) x sin(x) = 0. 10 / 61

Theorem (The second l Hôpital Rule: for type [ / ]) Suppose that f and g are differentiable on the interval (a, b) and that g (x) 0 there. Suppose also that (i) (ii) Then lim g(x) = ±, and x a+ lim x a+ f (x) g = L (where L is finite, or or ). (x) f (x) lim x a+ g(x) = L. Again, similar results hold for lim x b and for lim x c, and the cases a = and b = are allowed. 11 / 61

Example 4: Evaluate (a) lim x Solution: x 2 ; and (b) ex lim x ln(x). x 0+ 12 / 61

Example 4: Evaluate (a) lim x Solution: (a) By the second l Hôpital Rule, x 2 ; and (b) ex lim x ln(x). x 0+ x 2 lim x e x = lim 2x x e x = lim 2 x e x = 0. (b) Note that x ln(x) with x 0+ is of type [0 ]. To apply l Hôpital s Rules, we rewrite it as ln(x) x 1 so that ln(x) lim x ln(x) = lim x 0+ x 0+ x 1 = lim 1/x = lim ( x) = 0. x 0+ x 2 x 0+ 13 / 61

Do not try to use l Hôpital s Rules to evaluate limits that are not indeterminate of type [0/0] or [ / ]. Such attempts will almost always lead to false conclusions. No conclusion about the limit of f (x)/g(x) can be made using either l Hôpital s Rule if lim f (x)/g (x) does not exist. x 2 sin(1/x) One such example is lim = 0 by the Squeeze x 0 sin(x) theorem. However, lim [f (x)/g (x)] does not exist. In such x 0 cases, we need to use other techniques. The easiest way to deal with indeterminate forms of types [0 0 ], [ 0 ], and [1 ] is taking logarithms of the expressions involved. 14 / 61

Example 5: Evaluate lim x x. x 0+ Solution: Note that this indeterminate form is of type [0 0 ]. Let y = x x. Then by Example 4, This leads to lim ln(y) = lim x ln(x) = 0. x 0+ x 0+ lim x x = lim y = exp x 0+ x 0+ { } lim ln(y) x 0+ = e 0 = 1. 15 / 61

4.4 Extreme Values Maximum and Minimum Values Definition (Absolute Extreme Values) Function f has an absolute maximum value f (x 0 ) at the point x 0 in its domain if f (x) f (x 0 ) holds for every x in the domain of f. Similarly, f has an absolute minimum value f (x 1 ) at the point x 1 in its domain if f (x) f (x 1 ) holds for every x in the domain of f. 16 / 61

A function can have at most one absolute maximum or minimum value, although this value can be assumed at many points. Maximum and minimum values of a function are collectively referred to as extreme values. A function may not have any absolute extreme value. (i) f (x) = 1/x has no finite absolute maximum. (ii) g(x) = x with domain x (0, 1) has no extreme values. 17 / 61

Theorem (Existence of Extreme Values) If the domain of the function f is a closed, finite interval or a union of finitely many such intervals, and if f is continuous on that domain, then f must have an absolute maximum value and an absolute minimum value. Remark: This existence theorem is a restatement (with slight generalization) of the Max-Min Theorem in 1.4. 18 / 61

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Local Extreme Values Definition Function f has a local maximum value (loc max) f (x 0 ) at the point x 0 in its domain provided there exists a number h > 0 such that f (x) f (x 0 ) whenever x is in the domain of f and x x 0 < h. Similarly, f has a local minimum value (loc min) f (x 1 ) at the point x 1 in its domain provided there exists a number h > 0 such that f (x) f (x 1 ) whenever x is in the domain of f and x x 1 < h. 20 / 61

Critical Points, Singular Points, and Endpoints (i) critical points of f : points x D(f ) where f (x) = 0, (ii) singular points of f : points x D(f ) where f (x) is not defined, and (iii) endpoints of the domain of f : points x D(f ) that do not belong to any open interval contained in D(f ). Remark: In the previous figure, x 1, x 3, x 4 and x 6 are critical points, x 2 and x 5 are singular points, and a and b are endpoints. 21 / 61

Theorem (Locating Extreme Values) If the function f is defined on an interval I and has a local maximum (or local minimum) value at point x = x 0 in I, then x 0 must be either (i) a critical point of f, (ii) a singular point of f, or (iii) an endpoint of I. Finding Absolute Extreme Values: If f is defined on a closed interval or a union of finitely many closed intervals, we may find the absolute extreme value by checking the values of f at all critical, singular and endpoints. 22 / 61

Example 6: Find the maximum and minimum values of the function g(x) = x 3 3x 2 9x + 2 on the interval 2 x 2. Solution: Since g(x) is a polynomial, g (x) always exists so that it has no singular points. The derivative of g(x) is g (x) = 3x 2 6x 9 = 3(x + 1)(x 3). By letting g (x) = 0 we have x = 1 or x = 3. Since x = 3 is not in the domain, g has only one critical point at x = 1. Together with the two endpoints x = 2 and x = 2, we have g( 2) = 0, g( 1) = 7, g(2) = 20. Comparing the above values, we conclude that the maximum value of g(x) on 2 x 2 is 7 at x = 1 and the minimum value is 20 at x = 2. 23 / 61

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Example 7: Find the maximum and minimum values of h(x) = 3x 2/3 2x on the interval [ 1, 1]. Solution: The derivative of h(x) is h (x) = 2(x 1/3 1). Note that x 1/3 is not defined at x = 0, therefore x = 0 is a singular point of h. Also, h has a critical point where x 1/3 = 1, that is, at x = 1 (also an endpoint). Together with the other endpoint x = 1, we have h( 1) = 5, h(0) = 0, h(1) = 1. Comparing the above values, we conclude that the maximum value of h(x) on the interval [ 1, 1] is 5 at x = 1 and the minimum value is 0 at x = 0. 25 / 61

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4.5 Concavity and Inflections Definition We say that the function f is concave up (or convex) on an open interval I if it is differentiable there and the derivative f is an increasing function on I. Similarly, f is concave down on I if f exists and is an decreasing function on I. 27 / 61

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(i) If f is concave up on an interval, then, on that interval, the graph of f lies above its tangents, and chords joining points on the graph lie above the graph. (ii) If f is concave down on an interval, then, on that interval, the graph of f lies below its tangents, and chords to the graph lie below the graph. (iii) If the graph of f has a tangent at a point, and if the concavity of f is opposite on opposite sides of that point, then the graph crosses its tangent at that point (e.g., the point (b, f (b))). Such a point is called an inflection point of the graph of f. 29 / 61

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Theorem (Concavity and the Second Derivative) (a) If f (x) > 0 on interval I, then f is concave up on I. (b) If f (x) < 0 on interval I, then f is concave down on I. (c) If f has an inflection point at x 0 and f (x 0 ) exists, then f (x 0 ) = 0. 31 / 61

The Second Derivative Test Theorem The following rules determine whether the function has a local maximum or a local minimum value at a critical point x 0 : (a) If f (x 0 ) = 0 and f (x 0 ) < 0, then f has a local maximum value at x 0. (b) If f (x 0 ) = 0 and f (x 0 ) > 0, then f has a local minimum value at x 0. (c) If f (x 0 ) = 0 and f (x 0 ) = 0, no conclusion can be drawn. Specifically, f may have a local maximum at x 0, a local minimum, or an inflection point. 32 / 61

Example 8: Find and classify the critical points of f (x) = x 2 e x. Solution: The first derivative of f is f (x) = x(2 x)e x. Letting f (x) = 0, we know that f has two critical points: x = 0 and x = 2. To determine if they are local maximum or minimum values, we calculate the second derivative of f as f (x) = (2 4x + x 2 )e x. Note that f (0) = 2 > 0 and f (2) = 2e 2 < 0. We conclude that f has a local minimum value at x = 0 and a local maximum value at x = 2. 33 / 61

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4.6 Sketching the Graph of a Function Good-to-have skill. There are many problems in which being able to see the function from its formula helps you keep its general features in mind and makes solving the problem easier. When sketching the graph y = f (x), we have three sources of useful information: (i) the function f itself: determine the intercepts, components, symmetry, and asymptotes; (ii) the first derivative f : determine the intervals of increase and decrease and the location of extremes vales; (iii) the second derivative f : determine the concavity and inflection points, and sometimes extreme values. 35 / 61

Definition (1) The graph of y = f (x) has a vertical asymptote at x = a if either lim f (x) = ± or lim x a f (x) = ± or both. x a+ (2) The graph of y = f (x) has a horizontal asymptote y = L if either lim f (x) = L or lim x f (x) = L or both. x (3) The graph of y = f (x) has an oblique asymptote y = ax + b, where a 0, if either lim (f (x) (ax + b)) = 0 or lim x (f (x) (ax + b)) = 0 x or both. 36 / 61

Example 9: Find the asymptotes of the following functions: (a) f (x) = 1 x 2 x. Solution: The graph of f has a horizontal asymptote y = 0, and two vertical asymptotes at x = 0 and x = 1. 37 / 61

Example 9: Find the asymptotes of the following functions: (a) f (x) = 1 x 2 x. Solution: The graph of f has a horizontal asymptote y = 0, and two vertical asymptotes at x = 0 and x = 1. (b) g(x) = x 4 + x 2 x 4 + 1. Solution: The graph of g has a horizontal asymptote y = 1. 38 / 61

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Theorem (Asymptotes of Rational Functions) Suppose that f (x) = P m (x)/q n (x), where P m and Q n are polynomials of degrees of m and n, respectively. Suppose also that P m and Q n have no common linear factors. Then (a) The graph has a vertical asymptote at every position x such that Q n (x) = 0. (b) If m < n, the graph has a horizontal asymptote y = 0. (c) If m = n, the graph has a horizontal asymptote y = a m /b n. (d) If m = n + 1, the graph has an oblique asymptote y = ax + b, where a and b satisfy f (x) = ax + b + R(x)/Q n (x), the remainder R(x) is a polynomial of degree at most n 1. (e) If m > n + 1, the graph has no horizontal or oblique asymptotes. 43 / 61

Checklist for Curve Sketching: 1. Calculate f (x) and f (x), and express the results in factored form if possible. 2. Examine f (x) to determine its domain, the endpoints, the intercepts, the symmetry (Is f even or odd?), and the horizontal, vertical or oblique asymptotes. 3. Examine f (x) to determine the critical points, the singular points, and the intervals on which f is positive or negative. 4. Examine f (x) to determine the points where f (x) = 0, the points where f (x) is undefined, the intervals where f is positive (concave up) or negative (concave down), and the inflection points. 44 / 61

Example 10: Sketch the graph of f (x) = x 2 1 x 2 4. Solution: Note that f (x) = 6x (x 2 4) 2 and f (x) = 6(3x 2 + 4) (x 2 4) 3. From f : the domain is R except for { 2, 2}; the vertical asymptotes are x = 2 and x = 2; the horizontal asymptote is y = 1; f is symmetric on y-axis (i.e., f is even); the intercepts are ( 1, 0), (1, 0), an (0, 1/4); and other important points can be (3, 8/5) and ( 3, 8/5). From f : the critical point is x = 0; the singular points are x = 2 and x = 2; f (x) < 0 as x (0, 2) and x (2, ). From f : f (x) = 0 nowhere; f (x) < 0 as x (0, 2) and f (x) > 0 as x (2, ). 45 / 61

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Example 11: Sketch the graph of f (x) = xe x2 /2. Solution: Note that f (x) = (1 x 2 )e x2 /2 and f (x) = x(x 2 3)e x2 /2. From f : the domain is all x; the horizontal asymptote is y = 0, since 1 x 2 lim x ± e x2 /2 = lim 2x x ± xe x2 /2 = 0. f is symmetric about the origin (i.e., f is odd); the intercept is (0, 0) only, since e x2 /2 0 for all x. From f : the critical points are x = ±1 points (±1, ±1/ e) (±1, ±0.61); From f : f (x) = 0 at x = 0 and x = ± 3, points (0, 0), (± 3, 3e 3/2 ) (±1.73, ±0.39) 48 / 61

CP CP x 3 1 0 1 3 y 0 + + 0 y 0 + + 0 0 + y min max ) infl ( ( infl ) ) infl ( 49 / 61

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4.9 Linear Approximations Definition The linearization of the function f about a is the function L defined by L(x) = f (a) + f (a)(x a). We say that f (x) L(x) = f (a) + f (a)(x a) provides linear approximations for values of f near a. 51 / 61

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Example 12: Find linearizations of (a) f (x) = 1 + x about x = 0; and (b) g(t) = 1/t about t = 1/2. Solution: 53 / 61

Example 12: Find linearizations of (a) f (x) = 1 + x about x = 0; and (b) g(t) = 1/t about t = 1/2. Solution: (a) We have f (0) = 1 and f (0) = 1/(2 1 + x) x=0 = 1/2. The linearization of f about x = 0 is L(x) = f (0) + f (0)(x 0) = 1 + x 2. (b) We have g(1/2) = 2 and g (1/2) = 1/t 2 t=1/2 = 4. The linearization of g about t = 1/2 is L(t) = g( 1 2 ) + g ( 1 2 )(t 1 2 ) = 4 4t. 54 / 61

Example 13: Use the linearization for x about x = 25 to find an approximate value for 26. Solution: Let f (x) = x. We have f (25) = 5 and f (25) = 1 2 = 1 x x=25 10. The linearization of f about x = 25 is Putting x = 26, we get L(x) = f (25) + f (25)(x 25) = x 10 + 2.5. 26 26 = f (26) L(x) = + 2.5 = 5.1. 10 This is very close to the true value at 26 = 5.0990195... 55 / 61

4.10 Taylor Polynomials If we generalize the linear approximation to higher-degree approximation, it results in the Taylor polynomials. Let f (x) = c 0 + c 1 (x a) + c 2 (x a) 2 +, Assume that f (n) (a) exist for any positive integer n. We have f (n) (x) = c n n! + c n+1 (n + 1)! 1! (n + 2)! (x a) + c n+2 (x a) 2 + 2! Note that f (n) (a) = c n n!. We have c n = f (n) (a) n! n = 0, 1, 2, 56 / 61

Plugging in c n, we have f (x) = f (a) + f (a)(x a) + f (a) (x a) 2 2! + f (3) (a) (x a) 3 + + f (n) (a) (x a) n + Remainder 3! n! = P n (x) + Remainder. P n is called the nth-order Taylor polynomial for f about a. When a = 0, the Taylor polynomial P n becomes P n (x) = f (0) + f (0)x + f (0) x 2 + + f (n) (0) x n. 2! n! This is also referred to as the Maclaurin polynomial. 57 / 61

Some important Taylor polynomials: 1) e x = 1 + x + x 2 2! + x 3 3! + x 4 2) sin(x) = x x 3 3! + x 5 5! x 7 3) cos(x) = 1 x 2 2! + x 4 4! x 6 4! + 7! + 6! + 4) ln(1 + x) = x x 2 2 + x 3 3 x 4 + ( x < 1) 4 5) 1 1 + x = 1 x + x 2 x 3 + x 4 ( x < 1) 58 / 61

Example 14: Find an approximate value of e = 2.71828183 using the Taylor polynomials. Solution: Note that the Taylor series of function e x is Letting x = 1, we have e x = 1 + x + x 2 2! + x 3 3! + e = 1 + 1 + 1 2! + 1 3! + (1) If approximate e using the first 3 terms, we have e 2.5; (2) If approximate e using the first 4 terms, we have e 2.67; (3) If approximate e using the first 10 terms, we have e 2.71828. 59 / 61

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Example 15: Find the Taylor polynomial of f (x) = x sin(2x) about x = 0. Solution: Note that sin(x) = k=1 ( 1)k 1 1 (2k 1)! x 2k 1. We have sin(2x) = = Finally, ( 1) k 1 1 (2k 1)! (2x)2k 1 ( 1) k 1 2 2k 1 (2k 1)! x 2k 1. k=1 k=1 f (x) = x sin(2x) = ( 1) k 1 2 2k 1 (2k 1)! x 2k. k=1 61 / 61