Second Derivative Test

7 Key excerpts on "Second Derivative Test"

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  eBook - PDF

  Applied Calculus for the Managerial, Life, and Social Sciences

  A Brief Approach

  • Soo Tan(Author)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Also, f 0 1 4 2 5 6 1 4 2 1 2 5 18 . 0 and the Second Derivative Test implies that f 1 4 2 5 248 is a relative minimum of f, which confirms the results obtained earlier. Comparing the First and Second Derivative Tests Notice that both the First Derivative Test and the Second Derivative Test are used to classify the critical numbers of f. What are the pros and cons of the two tests? Since the Second Derivative Test is applicable only when f 0 exists, it is less versatile than the First Derivative Test. For example, it cannot be used to locate the relative mini- mum f 1 0 2 5 0 of the function f 1 x 2 5 x 2/3 . Furthermore, the Second Derivative Test is inconclusive when f 0 is equal to zero at a critical number of f, whereas the First Derivative Test always yields a conclusion; that is, it tells us if f has a relative maximum, relative minimum, or neither. The Sec- ond Derivative Test is also inconvenient to use when f 0 is difficult to compute. On the plus side, if f 0 is computed easily, then we use the Second Derivative Test, since it involves just the evaluation of f 0 at the critical number(s) of f. Also, the conclusions of the Second Derivative Test are important in theoretical work. We close this section by summarizing the different roles played by the first derivative f 9 and the second derivative f 0 of a function f in determining the properties of the graph of f. The first derivative f 9 tells us where f is increasing and where f is decreasing, whereas the second derivative f 0 tells us where the graph of f is concave upward and where it is concave downward. These different properties of f are reflected Explore and Discuss Suppose a function f has the following properties: 1. f s 1 x 2 . 0 for all x in an interval 1 a, b 2 . 2. There is a number c between a and b such that f r 1 c 2 5 0. What special property can you ascribe to the point 1 c, f 1 c 22 ? Answer the question if Property 1 is replaced by the property that f s 1 x 2 , 0 for all x in 1 a, b 2 .

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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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  eBook - ePub

  Painless Calculus

  • Barron's Educational Series, Christina Pawlowski-Polanish(Authors)
  • 2021(Publication Date)
  • Barrons Educational Services

    (Publisher)

  x = –7, –1, 4, 8.

  2.A function attains a relative maximum at a critical value where the sign of the derivative changes from positive to negative or, in this case, when the graph changes from above the x-axis to below the x-axis. This occurs when x = –1 and x = 8.

  3.A function is concave down when f″ (x) < 0. When using the graph of f′ (x), f″ (x) < 0, when the graph is decreasing. This occurs when x is in the intervals (–3, 2)∪(6, 10).

  Second Derivative Test

  The Second Derivative Test is another way to determine relative extrema for a function. It states that if a function f (x) is twice differentiable at x = a and if f′ (a) = 0 the following are true.

  •If f″ (a) > 0, then f (x) has a relative minimum at x = a.

  •If f″ (a) < 0, f (x) has a relative maximum at x = a.

  In other words, if there is a horizontal tangent at x = a and the curve is also concave up at this value, this point is a relative minimum. Similarly, if there is a horizontal tangent at x = a and the curve is also concave down at this value, this point is a relative maximum. A relative minimum and a relative maximum are shown in the figures below.

  The Second Derivative Test is painless and involves three steps.

  Step 1: Find the critical values of the function, the x-values where the first derivative is equal to zero.

  Step 2: Substitute the critical values into the second derivative of the function.

  Step 3: Determine the behavior of the point based on the sign of the second derivative.

  Example 28:

  Locate and describe the relative extrema of f (x) = x4 – 2x2 .

  Solution:

  One way to find the relative extrema is to use the Second Derivative Test.

  Step 1: Find the critical values of the function, the x-values where the first derivative is equal to zero:

  f′ (x) = 4x3 – 4x = 0

  4x(x2 – 1) = 0

  x = 0, x = – 1, x = 1

  Step 2:

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  Single Variable 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

  Single Variable

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

    (Publisher)

  b. What does the Second Derivative Test tell you about the nature of these stationary points? c. What does the first derivative test tell you about the nature of these stationary points? 6. a. Show that both f (x) = 1 − x 3 and g(x) = 2x 3 − x 6 have sta- tionary points at x = 0. b. What does the Second Derivative Test tell you about the nature of these stationary points? c. What does the first derivative test tell you about the nature of these stationary points? 7–14 Locate the critical points and identify which critical points are stationary points. 7. f (x) = 4x 4 − 16x 2 + 17 8. f (x) = 3x 4 + 12x 9. f (x) = x + 3 x 2 + 7 10. f (x) = x 2 + 8 x + 1 11. f (x) = 3 √ x 2 − 25 12. f (x) = (3x + 16) 2/3 (x 2 + 12) 13. f (x) = |sin(π x/2)| 14. f (x) = sin |π x/2| 15–18 True–False Assume that f is continuous everywhere. Deter- mine whether the statement is true or false. Explain your answer. 15. If f has a relative maximum at x = 1, then f (1) ≥ f (2). 16. If f has a relative maximum at x = 1, then x = 1 is a critical point for f . 17. If f  (x) > 0, then f has a relative minimum at x = 1. 18. If p(x) is a polynomial such that p  (x) has a simple root at x = 1, then p has a relative extremum at x = 1. Focus on Concepts 19–20 The graph of a function f (x) is given. Sketch graphs of y = f  (x) and y = f  (x). 19. 20. 1 2 y = f (x) 3 4 5 6 –1 x y 5 4 3 2 1 6 7 8 x y y = f (x) 21–24 Use the graph of f  shown in the figure to estimate all values of x at which f has (a) relative minima, (b) relative max- ima, and (c) inflection points. (d) Draw a rough sketch of the graph of a function f with the given derivative. 3.2 Analysis of Functions II: Relative Extrema; Graphing Polynomials 145 21. 22. x y 1 y = f ′(x) x y 2 y = f ′(x) 1 3 23. 24. x y -1 1 2 3 y = f ′(x) x y –1 1 2 3 4 5 y = f ′(x) 25–28 Use the given derivative to find all critical points of f, and at each critical point determine whether a relative maximum, relative minimum, or neither occurs.

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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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  Introduction to Differential Calculus

  Systematic Studies with Engineering Applications for Beginners

  • Ulrich L. Rohde, G. C. Jain, Ajay K. Poddar, A. K. Ghosh(Authors)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  Applications of derivatives 19a-Increasing and decreasing functions, and the sign of the first derivative horizontal tangents and local maximum/minimum values of functions Concavity, points of inflection, and the sign of the second derivative.

  Note (2): It must be noted that the notion of increasing and decreasing functions are always defined in terms of increasing x . Thus, as we move from left to right along the graph of a function,

  • in the case of an increasing function, the height of the graph continuously increases, and
  • in the case of a decreasing function, the height of the graph continuously decreases.

  19a.1.2

  In this chapter, we will use differentiation to find out whether a function is increasing or decreasing or neither.

  The derivative of a function y = f (x ) is the rate at which y changes with respect to x . It defines the slope of the function's graph at x and allows us to estimate how much y changes when we change x by a small amount. These concepts were discussed in Chapter 16. However, it is useful to revise the process of computing approximate changes in the value of a function y = f (x ) when the independent variable x is changed by an small amount. The following example makes it clear.

  Example (1): Consider the function,

  (1)

  so that we have,

  (2)

  Equation (2) tells us that at x = 1, y ′ = 3(1)2 = 3 and similarly at x =2, y ′ = 3(2)2 = 12.

  These calculations tell us that if we change the value of x by a small amount, say 0.2 units at x = 1, then the value of y [=x 3 ] will approximately change three times, that is, by 0.6 units. In other words, the height of the graph will be more (approximately) by 0.6 units at the point x = 1.2 than at the point x = 1. Similarly, the height of the graph at the point x = 2.2 will be more by 2.4 units (i.e., 12 times of 0.2) than at x = 2.0, and so on.

  19a.1.3

  If a function y = f (x ) has a derivative at a point x 0 , then we know that f is a continuous at x 0

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