First Derivative Test

8 Key excerpts on "First Derivative Test"

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  Calculus: An Applied Approach, Brief, International Metric Edition

  • Ron Larson(Author)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  For example, compare the graphs of f ( x ) = x 3 and g ( x ) = x 4 as shown in Figure 3.27. Both second derivatives are zero when x = 0, but only the graph of f has a point of inflection at x = 0. This shows that before concluding that a point of inflection exists at a value of x for which f uni2033 ( x ) = 0, you must test to be certain that the concavity actually changes at that point. 1 1 -1 1 -1 -1 1 -1 y x y x f ( x ) = x 3 g ( x ) = x 4 f uni2033 ( 0 ) = 0, and ( 0, 0 ) is a point of g uni2033 ( 0 ) = 0, but ( 0, 0 ) is not a point of inflection. inflection. FIGURE 3.27 ( ) 2 1 -1 -3 -5 -4 -3 -2 -1 2 y x 16 7 1 2 ( -1, -2) , f ( x ) = x 4 + x 3 -3 x 2 + 1 Two Points of Inflection FIGURE 3.26 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 Section 3.3 Concavity and the Second-Derivative Test 191 The Second-Derivative Test In addition to testing for concavity, the second derivative can be used to perform a simple test for relative minima and relative maxima. If f is a function such that f uni2032 ( c ) = 0 and the graph of f is concave upward at x = c , then f ( c ) is a relative minimum of f . Similarly, if f is a function such that f uni2032 ( c ) = 0 and the graph of f is concave downward at x = c , then f ( c ) is a relative maximum of f , as shown in Figure 3.28. x x y c downward Concave Concave Relative maximum f uni2033 ( c ) < 0 f uni2033 ( c ) > 0 c upward Relative minimum y FIGURE 3.28 Second-Derivative Test Let f uni2032 ( c ) = 0, and let f uni2033 exist on an open interval containing c . 1. If f uni2033 ( c ) > 0, then f ( c ) is a relative minimum. 2. If f uni2033 ( c ) < 0, then f ( c ) is a relative maximum. 3. If f uni2033 ( c ) = 0, then the test fails. In such cases, you can use the First-Derivative Test to determine whether f ( c ) is a relative minimum, a relative maximum, or neither.

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  Calculus: Early Transcendentals, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  For instance, part (a) is true because f 0s xd . 0 near c and so f is concave upward near c. This means that the graph of f lies above its horizontal tangent at c and so f has a local minimum at c. (See Figure 11.) NOTE The Second Derivative Test is inconclusive when f 0scd - 0. In other words, at such a point there might be a maximum, there might be a minimum, or there might be neither. This test also fails when f 0scd does not exist. In such cases the First Derivative Test must be used. In fact, even when both tests apply, the First Derivative Test is often the easier one to use. EXAMPLE 6 Discuss the curve y - x 4 2 4x 3 with respect to concavity, points of inflection, and local maxima and minima. SOLUTION If f s xd - x 4 2 4x 3 , then f 9 s xd - 4x 3 2 12x 2 - 4x 2 s x 2 3d f 0s xd - 12x 2 2 24x - 12x s x 2 2d To find the critical numbers we set f 9 s xd - 0 and obtain x - 0 and x - 3. (Note that f 9 is a polynomial and hence defined everywhere.) To use the Second Derivative Test we evaluate f 0 at these critical numbers: f 0s0d - 0 f 0s3d - 36 . 0 Since f 9 s3d - 0 and f 0s3d . 0, the Second Derivative Test tells us that f s3d - 227 is a local minimum. Because f 0s0d - 0, the Second Derivative Test gives no infor- mation about the critical number 0. But since f 9 s xd , 0 for x , 0 and also for 0 , x , 3, the First Derivative Test tells us that f does not have a local maximum or minimum at 0. x y=_2 0 1 2 _2 y FIGURE 10 f ª(c)=0 c P x y 0 f FIGURE 11 f 99scd . 0, f is concave upward Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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  Calculus

  Late Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  We are assuming Use the First Derivative Test to confirm the behavior at x 0 of each graph in Fig- ure 3.2.6. that f  (x) > 0 on the interval (a, x 0 ) and that f  (x) < 0 on the interval (x 0 , b), and we want to show that f (x 0 ) ≥ f (x) 142 Chapter 3 / The Derivative in Graphing and Applications for all x in the interval (a, b). However, the two hypotheses, together with Theorem 3.1.2 and its associated marginal note imply that f is increasing on the interval (a, x 0 ] and decreasing on the interval [x 0 , b). Thus, f (x 0 ) ≥ f (x) for all x in (a, b) with equality only at x 0 . Example 4 We showed in Example 3 that the function f (x) = 3x 5/3 − 15x 2/3 has critical points at x = 0 and x = 2. Figure 3.2.5 suggests that f has a relative maximum at x = 0 and a relative minimum at x = 2. Confirm this using the First Derivative Test. Solution. We showed in Example 3 that f  (x) = 5(x − 2) x 1/3 A sign analysis of this derivative is shown in Table 3.2.1. The sign of f  changes from + to − at x = 0, so there is a relative maximum at that point. The sign changes from − to + at x = 2, so there is a relative minimum at that point. Table 3.2.1 SECOND DERIVATIVE TEST There is another test for relative extrema that is based on the following geometric obser- vation: A function f has a relative maximum at a stationary point if the graph of f is concave down on an open interval containing that point, and it has a relative minimum if it is concave up (Figure 3.2.7). Figure 3.2.7 3.2.4 THEOREM (Second Derivative Test) Suppose that f is twice differentiable at the point x 0 . (a) If f  (x 0 ) = 0 and f  (x 0 ) > 0, then f has a relative minimum at x 0 . (b) If f  (x 0 ) = 0 and f  (x 0 ) < 0, then f has a relative maximum at x 0 . (c) If f  (x 0 ) = 0 and f  (x 0 ) = 0, then the test is inconclusive; that is, f may have a relative maximum, a relative minimum, or neither at x 0 . We will prove parts (a) and (c) and leave part (b) as an exercise. The second derivative test is often eas- ier to apply than the First Derivative Test. However, the First Derivative Test can be used at any critical point of a continuous function, while the second derivative test applies only at station- ary points where

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  Calculus Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  208 Chapter 4 / The Derivative in Graphing and Applications We will prove parts (a) and (c) and leave part (b) as an exercise. The second derivative test is often easier to apply than the First Derivative Test. However, the First Derivative Test can be used at any critical point of a continuous function, while the second derivative test applies only at stationary points where the second derivative exists. Proof (a) We are given that f ′ (x 0 ) = 0 and f ′′ (x 0 ) > 0, and we want to show that f has a relative minimum at x 0 . Expressing f ′′ (x 0 ) as a limit and using the two given conditions we obtain f ′′ (x 0 ) = lim x→x 0 f ′ (x) − f ′ (x 0 ) x − x 0 = lim x→x 0 f ′ (x) x − x 0 > 0 This implies that for x sufficiently close to but different from x 0 we have f ′ (x) x − x 0 > 0 (1) Thus, there is an open interval extending left from x 0 and an open interval extending right from x 0 on which (1) holds. On the open interval extending left the denominator in (1) is negative, so f ′ (x) < 0, and on the open interval extending right the denominator is positive, so f ′ (x) > 0. It now follows from part (b) of the First Derivative Test (Theorem 4.2.3) that f has a relative minimum at x 0 . ■ Proof (c) To prove this part of the theorem we need only provide functions for which f ′ (x 0 ) = 0 and f ′′ (x 0 ) = 0 at some point x 0 , but with one having a relative minimum at x 0 , one having a relative maximum at x 0 , and one having neither at x 0 . We leave it as an exercise for you to show that three such functions are f(x) = x 4 (relative minimum at x = 0), f(x) = −x 4 (relative maximum at x = 0), and f(x) = x 3 (neither a relative maximum nor a relative minimum at x 0 ). ■ – 2 –1 1 2 – 2 –1 1 2 x y y = 3x 5 – 5x 3 ▴ Figure 4.2.8 ▶ Example 5 Find the locations of the relative extrema of f(x) = 3x 5 − 5x 3 .

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  Use the First Derivative Test to confirm the behavior at x 0 of each graph in Figure 3.2.6. Theorem 3.2.3: First Derivative Test Suppose that f is continuous at a critical point x 0 . (a) If f  (x) > 0 on an open interval extending left from x 0 and f  (x) < 0 on an open interval extending right from x 0 , then f has a relative maximum at x 0 . (b) If f  (x) < 0 on an open interval extending left from x 0 and f  (x) > 0 on an open interval extending right from x 0 , then f has a relative minimum at x 0 . (c) If f  (x) has the same sign on an open interval extending left from x 0 as it does on an open interval extending right from x 0 , then f does not have a relative extremum at x 0 . 140 CHAPTER 3 The Derivative in Graphing and Applications Proof We will prove part (a) and leave parts (b) and (c) as exercises. We are assuming that f  (x) > 0 on the interval (a, x 0 ) and that f  (x) < 0 on the interval (x 0 , b), and we want to show that f (x 0 ) ≥ f (x) for all x in the interval (a, b).

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  Applied Calculus

  for Business, Life, and Social Sciences

  • Denny Burzynski(Author)
  • 2014(Publication Date)
  • XYZ Textbooks

    (Publisher)

  VIDEO EXAMPLES SECTION 3.1 Example 1 N(t) N(t) t 1 2 3 4 5 6 7 8 9 10 Figure 3 3.1 The First Derivative and the Behavior of Functions 217 Example 1 illustrates that graphs allow us to make estimates rather than obtain exact results. But very often in applied problems, estimates are perfectly acceptable. When exact results are desired, an analytic approach is needed. Using informa- tion obtained from the first derivative of a function can be the analytic approach needed to find the intervals upon which a function increases or decreases. Sign charts are useful for determining and displaying the intervals upon which a function increases or decreases. A sign chart is a number line picture on which the signs of the derivative of a function are displayed. Those signs indicate if the function is increasing or decreasing at a particular location. In constructing sign charts, we use the fact that the graph of a continuous func- tion possesses no holes, gaps, or breaks. A continuous function (in this case, the first derivative function) can change signs from positive to negative or from nega- tive to positive only if it equals zero or is undefined at some point. See Figure 4. Figure 4 shows the graph of the derivative of a continuous function f (x). Notice that the graph of the derivative function changes signs as it passes through the zero at x = 5. To the left of x = 5, the graph of f ′(x) lies below the x-axis so that it is negative (−). This tells us that the function f (x) itself is decreasing to the left of x = 5. To the right of x = 5, the graph of f ′(x) lies above the x-axis so that it is positive (+). This tells us that the function f (x) itself is increasing to the right of x = 5. Since the number 5 is critical to the accurate description of the behavior of f (x), it is called a critical value of f (x).

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  A Course of Mathematical Analysis

  International Series of Monographs on Pure and Applied Mathematics

  • A. F. Bermant, I. N. Sneddon, S. Ulam, M. Stark(Authors)
  • 2016(Publication Date)
  • Pergamon

    (Publisher)

  Whatever the circumstances, a point at which the derivative vanishes is called a stationary point of the function (its rate of change is zero). NECESSARY TEST FOR AN EXTREMUM. If a function f(x) is dif-ferentiable at a point x Q and has an extremum there, f r (xo) = 0. Proof. If f (x 0 ) were different from zero, the function y = f(x) would be either increasing or decreasing at x 0 , by what has been proved, i.e. it would not have an extremum there. FIG. 74 This theorem has a simple geometrical interpretation: the tangent (if it exists) at a point of the curve y = f(x) corresponding to an extremum must be parallel to Ox (Fig. 74). But a function may have an extremum at individual points at which it is non-differentiable (Fig. 74). Thus the extremal points of a function can only be points at which the derivative vanishes (i.e. stationary points) or does not exist. We can judge the behaviour of a function a t a point from its numerical derivative. To see how a function varies throughout a given interval, we have to turn to the derivative itself (the derived function). 2. Applications of the First Derivative 65. Theorems of Rolle and Lagrange. ROLLE'S THEOREM*. If the function f(x) is continuous in a closed interval [x l9 Λ^], is differentiable in the open interval (x 9 χ 2 ) and has * M. Rolle was a French mathematical! (1652—1719). 206 COURSE OF MATHEMATICAL ANALYSIS equal values at the ends of the interval, there exists at least one value x = ξ in this interval for which / ' ( § ) = 0. Proof. If the values of the function are equal at the ends of the interval, /(a^) = /(# 2 )> then either the function does not change at all inside the interval, and always f(x) = /(a^) = f(x 2 ), o r e ^ s e it does vary; if it varies, it attains a greatest (as in Fig. 75) or a least value, this value being evidently inside the interval [x l9 x 2 ], i.e. y< 0 M M, «1 * M 2 < 2 X FIG. 75 the function has an extremum inside the interval.

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  Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  Moreover, at the critical points in the first row the derivatives have opposite signs on the two sides of x 0 , whereas at the critical points in the second row the signs of the derivatives are the same on both sides. This suggests: Figure 4.2.5 A function f has a relative extremum at those critical points where f  changes sign. Figure 4.2.6 We can actually take this a step further. At the two relative maxima in Figure 4.2.6 the derivative is positive on the left side and negative on the right side, and at the two relative minima the derivative is negative on the left side and positive on the right side. All of this is summarized more precisely in the following theorem. 4.2.3 THEOREM (First Derivative Test) Suppose that f is continuous at a critical point x 0 . (a) If f  (x) > 0 on an open interval extending left from x 0 and f  (x) < 0 on an open interval extending right from x 0 , then f has a relative maximum at x 0 . (b) If f  (x) < 0 on an open interval extending left from x 0 and f  (x) > 0 on an open interval extending right from x 0 , then f has a relative minimum at x 0 . (c) If f  (x) has the same sign on an open interval extending left from x 0 as it does on an open interval extending right from x 0 , then f does not have a relative extremum at x 0 . Informally stated, parts (a) and (b) of Theorem 4.2.3 tell us that for a con- tinuous function, relative maxima occur at critical points where the derivative changes from + to − and relative min- ima where it changes from − to +. PROOF We will prove part (a) and leave parts (b) and (c) as exercises. We are assuming Use the First Derivative Test to confirm the behavior at x 0 of each graph in Fig- ure 4.2.6. that f  (x) > 0 on the interval (a, x 0 ) and that f  (x) < 0 on the interval (x 0 , b), and we want to show that f (x 0 ) ≥ f (x) for all x in the interval (a, b).

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