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PowerLecture_Chapter09 - 9 The Derivative Limits One-sided...

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Unformatted text preview: 9 The Derivative Limits One-sided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and Quotient Rules: The Higher-Order Derivatives The Chain Rule Differentiation of Exponential and Differentiation Logarithmic Functions Logarithmic Marginal Functions in Economics 9.1 9.1 Limits y 400 f ( x) 300 200 100 10 20 30 40 50 60 x Introduction to Calculus Introduction Historically, the development of calculus by Isaac Newton Historically, Isaac and Gottfried W. Leibniz resulted from the investigation Gottfried of the following problems: of 1. Finding the tangent line to a curve at a given point on the Finding tangent curve: curve: y T t Introduction to Calculus Introduction Historically, the development of calculus by Isaac Newton Historically, Isaac and Gottfried W. Leibniz resulted from the investigation Gottfried of the following problems: of 1. Finding the area of a planar region bounded by an Finding area arbitrary curve. arbitrary y R t Introduction to Calculus Introduction The study of the tangent-line problem led to the The tangent-line creation of differential calculus, which relies on the differential which concept of the derivative of a function. derivative The study of the area problem led to the creation of The area integral calculus, which relies on the concept of the integral which anti-derivative, or integral, of a function. anti-derivative or integral Example: A Speeding Maglev Example: From data obtained in a test run conducted on a From test prototype of maglev, which moves along a straight prototype which monorail track, engineers have determined that the position of the maglev (in feet) from the origin at time t position from time is given by is s = f(t) = 4t2 (0 ≤ t ≤ 30) 30) f is called the position function of the maglev. position The position of the maglev at time t = 0, 1, 2, 3, … , 10 is 0, The position f(0) = 0 (0) f(1) = 4 (1) f(2) = 16 (2) f(3) = 36 … f(10) = 400 (3) But what if we want to find the velocity of the maglev at But velocity any given point in time? point Example: A Speeding Maglev Example: Say we want to find the velocity of the maglev at t = 2. Say velocity We may compute the average velocity of the maglev We average over an interval of time, such as [2, 4] as follows: interval such [2, Distance covered f (4) − f (2) = Time elapsed 4−2 4(42 ) − 4(22 ) = 2 64 − 16 = 2 = 24 or 24 feet/second. or 24 This is not the velocity of the maglev at exactly t = 2, This not exactly but it is a useful approximation. approximation Example: A Speeding Maglev Example: We can find a better approximation by choosing a We better smaller interval to compute the speed, say [2, 3]. smaller [2, More generally, let t > 2. Then, the average velocity of let Then, average the maglev over the time interval [2, t] is given by time Distance covered f (t ) − f (2) = Time elapsed t−2 4(t 2 ) − 4(22 ) = t−2 4(t 2 − 4) = t−2 Example: A Speeding Maglev Example: 4(t 2 − 4) Average velocity = t−2 By choosing the values of t closer and closer to 2, we By closer we obtain average velocities of the maglev over smaller and average smaller time intervals. smaller The smaller the time interval, the closer the average The smaller time the closer velocity becomes to the instantaneous velocity of the train instantaneous at t = 2, as the table below demonstrates: t 2.5 2.1 2.01 2.001 2.0001 Average Velocity 18 16.4 16.04 16.004 16.0004 The closer t gets to 2, the closer the average velocity gets The closer the closer average to 16 feet/second. 16 Thus, the instantaneous velocity at t = 2 seems to be Thus, instantaneous 16 feet/second. 16 Intuitive Definition of a Limit Intuitive Consider the function g, which gives the average velocity Consider g, average of the maglev: of 4(t 2 − 4) g (t ) = t −2 Suppose we want to find the value that g(t) approaches Suppose approaches as t approaches 2. approaches ✦ We take values of t approaching 2 from the right We approaching from (as we did before), and we find that g(t) approaches approaches 16: 16 t 2.5 2.1 2.01 2.001 2.0001 g (t ) 18 16.4 16.04 16.004 16.0004 ✦ Similarly, we take values of t approaching 2 from the left, Similarly, approaching from and we find that g(t) also approaches 16: approaches 16 t 1.5 1.9 1.99 1.999 1.9999 g(t) 14 15.6 15.96 15.996 15.9996 Intuitive Definition of a Limit Intuitive We have found that as t approaches 2 from either side, We approaches from g(t) approaches 16. approaches 16 In this situation, we say that the limit of g(t) as t In the approaches 2 is 16. approaches 16 This is written as 4(t 2 − 4) lim g (t ) = lim = 16 t →2 t →2 t −2 Observe that t = 2 is not in the domain of g(t) . Observe not But this does not matter, since t = 2 does not play any But does since role in computing this limit. role Limit of a Function Limit The function f has a limit L as x approaches a, written The function limit approaches lim f ( x ) = L x→a if the value of f(x) can be made as close to the number if close L as we please by taking x values sufficiently close to values sufficiently (but not equal to) a. not Examples Examples Let f(x) = x3. Evaluate Let Evaluate lim f ( x ). x→2 Solution Solution You can see in the graph You that f(x) can be as close can close to 8 as we please by taking x sufficiently close to 2. sufficiently Therefore, Therefore, f( x ) = x y 8 3 6 4 2 lim x 3 = 8 x→2 –2 –1 1 –2 2 3 x Examples Examples x + 2 1 Let g ( x ) = Let if x ≠ 1 if x = 1 Evaluate lim g ( x ). x→1 Solution Solution You can see in the graph that You g(x) can be as close can close to 3 as we please by taking x sufficiently close to 1. sufficiently Therefore, Therefore, y g(x) 5 3 1 lim g ( x ) = 3 x→1 –2 –1 1 2 3 x Examples Examples Let f ( x ) = Let 1 x2 Evaluate lim f ( x ). Solution Solution The graph shows us that as x The approaches 0 from either approaches side, f(x) increases without side bound and thus does not bound not approach any specific real number. number. Thus, the limit of f(x) does Thus, not exist as x approaches 0. not approaches x→0 y 5 f ( x) = –2 –1 1 2 1 x2 x Theorem 1 Theorem Properties of Limits lim f ( x ) = L x→a Suppose Then, 1. lim [ x→a and lim g ( x ) = M x→a r f ( x )] = lim f ( x ) = Lr x →a r r, a real number 2. lim cf ( x ) = c lim f ( x ) = cL 2. x→a x →a 3. lim x →a [ f ( x ) ± g ( x )] = lim x→a 4. lim 4. x→a [ f ( x ) g ( x )] = lim x→a 5. 5. f ( x ) ± lim g ( x ) = L ± M x→a f ( x ) lim g ( x ) = LM x→a f ( x ) lim f ( x ) L lim = x→a = x →a g ( x ) lim g ( x ) M x→a c, a real number Provided that M ≠ 0 Provided Examples Examples Use theorem 1 to evaluate the following limits: Use theorem 3 lim x = lim x = 23 = 8 x→2 3 x→2 3/2 lim 5x = 5 lim x = 5(4)3/2 = 40 x→4 3/2 x→4 4 lim (5x − 2) = 5 lim x − lim 2 = 5(1)4 − 2 = 3 x→1 4 x→1 x→1 Examples Examples Use theorem 1 to evaluate the following limits: Use theorem lim 2x x →3 x + 7 = 2 lim x x→3 3 2 3 lim x 2 + 7 = 2(3)3 (3)2 + 7 = 216 x→3 lim (2 x 2 + 1) 2(2)2 + 1 9 2x + 1 = x→2 = = =3 lim x→2 lim ( x + 1) 2 +1 3 x +1 x→2 2 Indeterminate Forms Indeterminate 4( x 2 − 4) Let’s consider lim x →2 x − 2 which we evaluated earlier for the maglev example by which maglev looking at values for x near x = 2. near If we attempt to evaluate this expression by applying If Property 5 of limits, we get Property lim 4( x 2 − 4) 0 4( x − 4) x→2 lim = = x →2 x − 2 lim x − 2 0 x →2 2 In this case we say that the limit of the quotient f(x)/g(x) as In quotient x approaches 2 has the indeterminate form 0/0. approaches 0/0 This expression does not provide us with a solution to our This not solution problem. problem. Strategy for Evaluating Indeterminate Forms Strategy 1. Replace the given function with an appropriate one that takes on the same values as the same original function everywhere except at x = a. everywhere except 2. Evaluate the limit of this function as x approaches a. approaches Examples Examples 4( x 2 − 4) Evaluate lim x →2 x − 2 Solution As we’ve seen, here we have an indeterminate form 0/0. As indeterminate 0/0 We can rewrite 4( x − 2)( x + 2) We 4( x 2 − 4) = = 4( x + 2) x−2 x−2 x≠2 Thus, we can x 2 − 4) 4( say that lim x →2 x−2 = lim 4( x + 2) = 16 x →2 Note that 16 is the same value we obtained for the maglev Note 16 example through approximation. example approximation Examples Examples 4( x 2 − 4) Evaluate lim x →2 x − 2 Solution Notice in the graphs below that the two functions yield the Notice same graphs, except for the value x = 2: same except 4( x 2 − 4) f ( x) = x−2 y 20 20 16 8 4 –1 12 8 –2 16 12 –3 g ( x ) = 4( x + 2) y 4 1 2 3 x –3 –2 –1 1 2 3 x Examples Examples Evaluate 1+ h −1 lim h→0 h Solution As we’ve seen, here we have an indeterminate form 0/0. As indeterminate 0/0 We can rewrite (with the constraint that h ≠ 0): We constraint 1+ h −1 1+ h −1 1+ h +1 h 1 = ⋅ = = h h 1 + h + 1 h( 1 + h + 1) 1+ h +1 Thus, we can say that 1+ h −1 1 1 1 lim = lim = = h→0 h→0 1 + h + 1 h 1 +1 2 Limits at Infinity Limits There are occasions when we want to know whether f(x) There approaches a unique number as x increases without approaches unique bound. bound In the graph below, as x increases without bound, f(x) In without approaches the number 400. 400 We call the line y = 400 We y a horizontal asymptote. horizontal 400 In this case, we can say In f ( x) that that 300 lim f ( x ) = 400 x→∞ and we call this a limit and limit of a function at infinity. 200 100 10 20 30 40 50 60 x Example Example 2x2 Consider the function f ( x ) = 1 + x2 Determine what happens to f(x) as x gets larger and larger. Determine larger Solution We can pick a sequence of values of x and substitute them We sequence in the function to obtain the following values: in x 1 2 5 10 100 1000 f(x) 1 1.6 1.92 1.98 1.9998 1.999998 As x gets larger and larger, f(x) gets closer and closer to 2. As larger closer Thus, we can say that 2x2 lim x→∞ 1 + x 2 =2 Limit of a Function at Infinity Limit The function f has the limit L as x increases without The limit bound (as x approaches infinity), written bound approaches lim f ( x ) = L x→∞ if f(x) can be made arbitrarily close to L by taking x if close large enough. large Similarly, the function f has the limit M as x decreases Similarly, limit without bound (as x approaches negative infinity), without approaches ), written written lim f ( x ) = M x→−∞ if f(x) can be made arbitrarily close to M by taking x if close large enough in absolute value. large Examples Examples −1 Let f ( x ) = Let 1 Evaluate if x < 0 if x ≥ 0 lim f ( x ) lim f ( x ) and x→∞ x→−∞ Solution Graphing f(x) reveals that Graphing y lim f ( x ) = 1 x→∞ lim f ( x ) = −1 x→−∞ f ( x) 1 –3 3 –1 x Examples Examples Let g ( x ) = Let Evaluate 1 x2 lim g ( x ) and lim g ( x ) x→∞ x→−∞ Solution Graphing g(x) reveals that Graphing y 1 g ( x) = 2 x lim g ( x ) = 0 x→∞ lim g ( x ) = 0 x→−∞ –3 –2 –1 1 2 3 x Theorem 2 Theorem Properties of Limits All properties of limits listed in Theorem 1 are valid when a is replaced by ∞ or –∞ . In addition, we have the following properties for limits to infinity: For all n > 0, For 1 lim =0 x→∞ x n and 1 provided that n is defined. x 1 lim =0 x→−∞ x n Examples Examples x2 − x + 3 Evaluate lim Evaluate x→∞ 2 x3 + 1 Solution Solution The limits of both the numerator and denominator do not The limits numerator denominator exist as x approaches infinity, so property 5 is not exist property not applicable. applicable. We can find the solution instead by dividing numerator We dividing and denominator by x3: and by 11 3 − 2+ 3 ( x 2 − x + 3) / x 3 xx x = 0−0+0 = 0 = 0 lim = lim x→∞ (2 x 3 + 1) / x 3 x→∞ 1 2+0 2 2+ 3 x Examples Examples 3x 2 + 8 x − 4 Evaluate lim Evaluate x→∞ 2 x2 + 4 x − 5 Solution Solution Again, we see that property 5 does not apply. Again, property not So we divide numerator and denominator by x2: So divide and by 84 −2 (3x + 8 x − 4) / x x x = 3+ 0−0 = 3 lim 2 = lim x→∞ (2 x + 4 x − 5) / x 2 x→∞ 45 2+ − 2 2+0−0 2 xx 2 2 3+ Examples Examples 2 x 3 − 3x 2 + 1 Evaluate lim Evaluate x→∞ 2 x + 2x + 4 Solution Solution Again, we see that property 5 does not apply. Again, property not But dividing numerator and denominator by x2 does not But dividing and by not 1 help in this case: help 2x − 3 + 2 3 2 2 lim x→∞ (2 x − 3x + 1) / x x = lim x→∞ 24 ( x 2 + 2 x + 4) / x 2 1+ + 2 xx In other words, the limit does not exist. In the We indicate this by writing 2 3 lim x→∞ 2 x − 3x + 1 =∞ 2 x + 2x + 4 9.2 9.2 One-Sided Limits and Continuity y a b c d x One-Sided Limits One-Sided x −1 x +1 Consider the function f ( x ) = if x < 0 if x ≥ 0 Its graph shows that Its f does not have a does not limit as x approaches limit zero, because zero because approaching from each side results in different values. values y f ( x) 1 –1 1 –1 x One-Sided Limits One-Sided x −1 x +1 Consider the function f ( x ) = if x < 0 if x ≥ 0 If we restrict x to be greater If restrict greater than zero (to the right of zero), than we see that f(x) approaches 1 as approaches close as we please as x approaches 0. approaches In this case we say that the In right-hand limit of f as x right-hand approaches 0 is 1, written approaches lim f ( x ) = 1 x→0+ y f ( x) 1 –1 1 –1 x One-Sided Limits One-Sided x −1 x +1 Consider the function f ( x ) = if x < 0 if x ≥ 0 Similarly, if we restrict x to be Similarly, restrict less than zero (to the left of zero), less we see that f(x) approaches –1 approaches –1 as close as we please as x approaches 0. approaches In this case we say that the In left-hand limit of f as x left-hand approaches 0 is – 1, written approaches lim f ( x ) = −1 x→0− y f ( x) 1 –1 1 –1 x One-Sided Limits One-Sided The function f has the right-hand limit L as x The right-hand approaches from the right, written approaches lim f ( x ) = L x→a + if the values of f(x) can be made as close to L as we if please by taking x sufficiently close to (but not equal to) a and to the right of a. to Similarly, the function f has the left-hand limit L as x Similarly, left-hand approaches from the left, written approaches lim f ( x ) = L x→a − if the values of f(x) can be made as close to L as we if please by taking x sufficiently close to (but not equal to) a and to the left of a. to Theorem 3 Theorem Properties of Limits The connection between one-sided limit and the two-sided limit defined earlier is given by the following theorem. Let f be a function that is defined for all values of x Let close to x = a with the possible exception of a itself. Then Then lim f ( x ) = L x →a if and only if lim f ( x ) = lim f ( x ) = L x→a − x→a + Examples Show that lim f ( x ) exists by studying the one-sided Show one-sided x →0 limits of f as x approaches 0: limits approaches x f ( x) = −x Solution Solution For x > 0, we find For if x > 0 if x ≤ 0 y lim f ( x ) = 0 x→0+ And for x ≤ 0, we find And lim f ( x ) = 0 1 x→0− Thus, Thus, lim f ( x ) = 0 x →0 f ( x) 2 –2 –1 1 2 x Examples Examples Show that lim g ( x ) does not exist. Show x →0 −1 g ( x) = 1 if x < 0 if x ≥ 0 Solution For x < 0, we find For y lim g ( x) = −1 x →0− And for x ≥ 0, we find And lim g ( x) = 1 g ( x) 1 x →0+ Thus, Thus, x lim g ( x) does x →0 not exist. not –1 Continuous Functions Continuous Loosely speaking, a function is continuous at a given point if Loosely function its graph at that point has no holes, gaps, jumps, or breaks. no holes gaps jumps or breaks Consider, for example, the graph of f Consider, y a x This function is discontinuous at the following points: This discontinuous ✦ At x = a, f is not defined (x = a is not in the domain of f ). At not not ). Continuous Functions Continuous Loosely speaking, a function is continuous at a given point if Loosely function its graph at that point has no holes, gaps, jumps, or breaks. no holes gaps jumps or breaks Consider, for example, the graph of f Consider, y a b x This function is discontinuous at the following points: This discontinuous ✦ At x = b, f(b) is not equal to the limit of f(x) as x approaches b. At not of as approaches Continuous Functions Continuous Loosely speaking, a function is continuous at a given point if Loosely function its graph at that point has no holes, gaps, jumps, or breaks. no holes gaps jumps or breaks Consider, for example, the graph of f Consider, y b c x This function is discontinuous at the following points: This discontinuous ✦ At x = c, the function does not have a limit, since the left-hand At does since left-hand and right-hand limits are not equal. right-hand not Continuous Functions Continuous Loosely speaking, a function is continuous at a given point if Loosely function its graph at that point has no holes, gaps, jumps, or breaks. no holes gaps jumps or breaks Consider, for example, the graph of f Consider, y c d x This function is discontinuous at the following points: This discontinuous ✦ At x = d, the limit of the function does not exist, resulting in a At limit resulting break in the graph. break Continuity of a Function at a Number Continuity A function f is continuous at a number x = a function continuous if the following conditions are satisfied: conditions 1. f(a) is defined. 2. lim f ( x ) exists. x→a 3. lim f ( x ) = f (a ) 3. x→a If f is not continuous at x = a, then f is said to If then be discontinuous at x = a. discontinuous Also, f is continuous on an interval if f is Also, continuous continuous at every number in the interval. continuous Examples Examples Find the values of x for which the function is continuous: Find continuous f ( x) = x + 2 Solution The function f is continuous everywhere because the three The continuous conditions for continuity are satisfied for all values of x. y 5 f ( x) = x + 2 4 3 2 1 – 2 –1 1 2 x Examples Examples Find the values of x for which the function is continuous: Find continuous x2 − 4 g ( x) = x−2 Solution The function g is discontinuous at x = 2 because g is not The defined at that number. It is continuous everywhere else. defined y x2 − 4 g ( x) = x−2 5 4 3 2 1 – 2 –1 1 2 x Examples Examples Find the values of x for which the function is continuous: Find continuous x + 2 h( x ) = 1 if x ≠ 2 if x = 2 Solution The function h is continuous everywhere except at x = 2 The continuous except where it is discontinuous because because h (2) = 1 ≠ lim h ( x ) = 4 x→2 5 y y = h( x ) 4 3 2 1 – 2 –1 1 2 x Examples Examples Find the values of x for which the function is continuous: Find continuous −1 if x < 0 F ( x) = 1 if x ≥ 0 Solution The function F is discontinuous at x = 0 because the limit The at of F fails to exist as x approaches 0. It is continuous fails approaches It everywhere else. y y = F ( x) 1 x –1 Examples Examples Find the values of x for which the function is continuous: Find continuous 1 if x > 0 G( x) = x −1 i f x ≤ 0 Solution The function G is discontinuous at x = 0 because the limit The at of G fails to exist as x approaches 0. It is continuous fails approaches It everywhere else. y y = G ( x) x –1 Properties of Continuous Functions Properties 1. The constant function f(x) = c is continuous everywhere. The constant continuous 2. The identity function f(x) = x is continuous everywhere. The identity continuous If f and g are continuous at x = a, then and are at 1. [f(x)]n, where n is a real number, is continuous at where continuous x = a whenever it is defined at that number. 2. f ± g is continuous at x = a. continuous 3. fg is continuous at x = a. fg continuous 4. f /g is continuous if g(a) ≠ 0. /g continuous Properties of Continuous Functions Properties Using these properties, we can obtain the following Using additional properties. additional 1. A polynomial function y = P(x) is continuous at every polynomial value of x. 2. A rational function R(x) = p(x)/q(x) is continuous at rational every value of x where q(x) ≠ 0. Examples Examples Find the values of x for which the function is continuous. Find values continuous f ( x ) = 3x 3 + 2 x 2 − x + 10 Solution The function f is a polynomial function of degree 3, so f(x) The polynomial so is continuous for all value...
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