First Derivative Test

5 Key excerpts on "First Derivative Test"

• 

  eBook - ePub

  CLEP® Calculus Book + Online

  • Gregory Hill, Mel Friedman(Authors)
  • 2013(Publication Date)
  • Research & Education Association

    (Publisher)

  Besides being used to determine intervals of concavity and inflection points, the second derivative can also be applied efficiently to justify the existence of a local extreme value on a function. Recall that the first-derivative test for maxima and minima required identifying critical points and checking whether the first derivative changed signs on either side of the critical point. This requires knowledge about how the function is behaving around the critical point. The second-derivative test is a local test applied right at the critical point.

  Second Derivative Test for Local Extremes

  Let c be a point on the interior of the domain of a function, f.

  If f′(c) = 0 and f″(c) > 0, then f has a local minimum at x = c.

  If f′(c) = 0 and f″(c) < 0, then f has a local maximum at x = c.

  If f′(c) = 0, f has a horizontal tangent at x = c. But if f′(c) > 0, then f is concave up, so the graph must lie above the horizontal tangent at c. Therefore, f must have a local minimum at x = c. Similar logic can be applied for a local maximum.

  EXAMPLE 5.14

  Use the second-derivative test to determine the x-coordinates of the local extreme values on f(x) = x3 + 3x2 – 9x + 2.

  SOLUTION

  f′(x) = 3x2 + 6x – 9, and f″(x) = 6x + 6.

  Solving f′(x) = 0 gives 3(x2 + 2x – 3) or 3(x + 3) (x – 1) = 0. So the critical values are x = –3 and x = 1.

  f″(–3) = –12, and f″(l) = 12, so f″(–3) < 0 and f″(1) > 0.

  By the second-derivative test, f has a local maximum at x = – 3 and a local minimum at x = 1.

  CURVE SKETCHING

  Mastering the relationships between a function and its first and second derivatives takes a good amount of practice. One way to test your conceptual understanding of these relationships is to work graphically with the three functions. The goal is to be able to synthesize information in such a way as to move seamlessly between a function and its derivatives. Given information about first and second derivatives, you ought to be able to make a rough sketch of the function. Given a function, you ought to be able to sketch the graph of its derivative.

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  eBook - ePub

  Mathematical Economics

  • Arsen Melkumian(Author)
  • 2012(Publication Date)
  • Routledge

    (Publisher)

  x) in the neighborhood of the stationary point. The second-derivative test below is probably the most popular test for local extrema.

  THEOREM 5.2: (SECOND -DERIVATIVE TEST FOR LOCAL EXTREMA ). Suppose that f (x) is twice differentiable on an open interval I and it has only one stationary point A = (a, f (a)) on that interval.

  (a) If f″(a) < 0, then A is a local maximum.

  (b) If f″(a) > 0, then A is a local minimum.

  (c) If f″(a) = 0, then A may be a local extremum or an inflection point.

  As we can see from Theorem 5.2, sometimes the second derivative evaluated at a critical point may vanish. In such a case we can use the higher-order derivative test for local extrema to classify our stationary points.

  THEOREM 5.3: (HIGHER -ORDER DERIVATIVE TEST FOR LOCAL EXTREMA ). Let f be a differentiable function on the interval I and let a ∈ I be a critical point of f (x) such that

      (i) f(1) (a) = f(2) (a) = ··· = f(n−1) (a) = 0

     (ii) f

  (n)

  (a) exists and is non-zero.

  (a) If n is even and f(n) (a) < 0, then (a, f (a)) is a local maximum.

  (b) If n is even and f(n) (a)> 0, then (a, f (a)) is a local minimum.

  (c) If n is odd, then (a, f (a)) is an inflection point.

  Note that the higher-order derivative test will work only if the function f (x) yields a nonzero derivative value f

  (n)

  (a) for some finite number n. While most functions will meet this condition, there are some rare functions that produce a zero-valued derivative f

  (n)

  (a

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  eBook - ePub

  Quantitative Methods for Business and Economics

  • Adil H. Mouhammed(Author)
  • 2015(Publication Date)
  • Routledge

    (Publisher)

  2 = f″(x). After finding the second derivative, substitute the critical point(s) we found in step 2 in the second derivative to find whether the second derivative is positive or negative. If it is positive, the function has a minimum value at the critical point(s). In contrast, if the second derivative is negative, the function has a maximum value at the critical point(s). Otherwise, the test fails and the function may have an inflection point.

  Example 1: Optimize the function y = 3x2 - 12x.

  Solution: Find the first derivative and set it equal to zero. That is, f′(x) = dy/dx = 6x - 12 = 0

  and solving for the critical point xc yields

  6x = 12

  and xc = 2. Now, find the second derivative for the function

  d2 y/dx2 = f″(x) = 6.

  Because the second derivative is positive, the function has a minimum value at the critical point xc = 2. Also, if we substitute various values of x in our function, the y’s values (or the values of the function) will be

  As can be seen, the function has a minimum value of (-12) at the critical point xc = 2. One should note from this example that we can find the minimum value of the function by using the first derivative only. As we know the critical value, which is xc = 2, we can take other values for x, such as x = 1 (which is less than the critical value) and x = 3 (which is higher than the critical value), and after substituting these values in the first derivative we can obtain f′(x) = - 6 and f′(x) = 6. Because the first derivative changes signs from negative to positive, the function must have a minimum value at the critical point.

  Example 2: Optimize the function y = f(x) = -2x2 + 20x - 2.

  Solution: The first derivative is

  dy/dx = -4x + 20 = 0, and hence xc = 5.

  The second derivative is

  d2 y/dx2 = - 4.

  As the second derivative is negative, the function has a maximum value at the critical point xc = 5. One should note that by substituting various values of x in the given function, we can find the maximum value of the function. This is shown below:

  Similarly, by using the first derivative alone, one can determine the maximum value of the function. Take two points such as x = 4, which is less than the critical point, and x = 6, which is higher than the critical point, and use these two points to evaluate the first derivative. Having done so, we obtain f′(x) = 4 and f′(x) = - 4. Because the first derivative changes signs from positive to negative, the function must have a maximum value at the critical point.

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  eBook - ePub

  Introductory Mathematical Economics

  • Adil H. Mouhammed(Author)
  • 2020(Publication Date)
  • Routledge

    (Publisher)

  2 = f ″(x). After finding the second derivative, substitute the critical point(s) we found in step 2 in the second derivative to find whether the second derivative is positive or negative. If it is positive, the function has a minimum value at the critical point(s). In contrast, if the second derivative is negative, the function has a maximum value at the critical point(s). Otherwise, the test fails and the function may have an inflection point.

  Example 1: Optimize the function y = 3x2 - 12x.

  Solution: Find the first derivative and set it equal to zero. That is,

  f ′

  ( x )

  = dy/dx = 6 x − 12 = 0

  and solving for the critical point xc yields

  6x = 12

  and xc = 2. Now, find the second derivative for the function

  d 2

  y/dx 2

  =

  f ?

  ( x )

  = 6.

  Because the second derivative is positive, the function has a minimum value at the critical point xc = 2. Also, if we substitute various values of x in our function, the y’s values (or the values of the function) will be

  x

  − 2

  − 1

  0 1 2 3

  y 36 15 0

  − 9

  − 12

  − 9

  As can be seen, the function has a minimum value of (−12) at the critical point xc = 2. One should note from this example that we can find the minimum value of the function by using the first derivative only. As we know the critical value, which is xc = 2, we can take other values for x, such as x = 1 (which is less than the critical value) and x = 3 (which is higher than the critical value), and after substituting these values in the first derivative we can obtain f ′(x) = - 6 and f ′(x) = 6. Because the first derivative changes signs from negative to positive, the function must have a minimum value at the critical point.

  Example 2: Optimize the function y = f(x) = −2x2 + 20x - 2.

  Solution: The first derivative is

  dy/dx = − 4 x + 20 = 0 ,

  and hence x c

  = 5.

  The second derivative is

  d 2

  y/dx 2

  = − 4.

  As the second derivative is negative, the function has a maximum value at the critical point xc

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  eBook - ePub

  AP&reg; Calculus AB/BC All Access Book + Online + Mobile

  • Stu Schwartz(Author)
  • 2013(Publication Date)
  • Research & Education Association

    (Publisher)

  second derivative test.

  If f′(k) = 0 and f″(k) > 0, f(x) has a relative minimum at x = k.

  If f′(k) = 0 and f″(k) < 0, f(x) has a relative minimum at x = k.

  If f′(k) = 0 and f″(k) = 0, the test is inconclusive.

   EXAMPLE 20:

  Given .

            a) Find the x-coordinate of all critical points of f and determine whether each is a relative minimum or relative maximum.

       b) Find intervals where the graph of f is concave up.

       c) Find points of inflection. Justify all answers.

   SOLUTIONS:

  Also, x = 0 is a critical value but the function is undefined there.

  f has a relative maximum at because f′ switches from positive to negative.

  f has a relative minimum at because f′ switches from negative to positive.

  f is concave up on and because f″(x) > 0.

  Note that the answer is incorrect as x = 0 is not in the domain of f.

            (c)  There are inflection points at only, as f″(x) changes sign at these values.

  DIDYOU KNOW?

  Inflection points have physical applications. If you are driving along a winding road, the inflection point is the location at which the steering-wheel is momentarily straight as you turn from right to left or vice versa. If you are running up a hill, the inflection point is the location on the hill which has maximum steepness. Running will be progressively harder until you reach the inflection point and then progressively easier.

   EXAMPLE 21 :

  The figure below shows the graph of y = f′(x), the derivative of the function f for −4 ≤ x ≤ 6. The graph of f′ has horizontal tangents at x = −2.5, x = −1.5, and x = 2.

  a) Find all values of x, −4 ≤ x ≤ 6, where f attains a relative maximum.

  b) Find all values of x, −4 ≤ x ≤ 6, where f has inflection points. Justify your answers.

   SOLUTIONS:

            (a)   f has a relative maximum at x = −2 only because f′ switches from positive to negative.

            (b) f has inflection pts. at x = −3, x = −2.5, x = −1.5, x = −1, and x = 2 because f

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