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Nature of Stationary Points

Using the first and second derivatives for a given function, we can identify the nature of stationary points for that function. Depending on the given function, we can get three types of stationary points:

  1. If f'(x) = 0 and f”(x) > 0, then there is a minimum turning point
  2. If f'(x) = 0 and f”(x) < 0, then there is a maximum turning point
  3. If f'(x) = 0 and f”(x) = 0, then there is a horizontal point of inflection provided there is a change in concavity

Here are a few examples to find the types and nature of the stationary points.

  1. Consider the curve f(x) = 3x4 – 4x3 – 12x2 + 1f'(x) = 12x3 – 12x2 – 24x

    = 12x(x2 – x – 2)

    For stationary point, f'(x) = 0

    Therefore 12x(x2 – x – 2) = 0

    doubleright  x = 0  or  x2 – x – 2 = 0

    x2 – x – 2 = 0

    x2 – 2x + x – 2 = 0

    x(x – 2)  + 1(x – 2)  = 0

    (x – 2)(x + 1)  =  0

    So we’ll have a stationary point at –  x = 0, x = -1 or x = 2.

    To find the coordinates of the stationary points, we apply the values of x in the equation.

    When x = 0, y = 3(0)4 – 4(0)3 – 12(0)2 + 1  =  1

    So (0, 1) is the first stationary point

    When x = -1, y = 3(-1)4 – 4(-1)3 – 12(-1)2 + 1

    =  3 – 4 -12 + 1  =  -4

    So (-1, -4) is the second stationary point

    When x = 2, y = 3(2)4 – 4(2)3 – 12(2)2 + 1

    =  48 – 32 -48 + 1  =  -31

    So (2, -31) is the third stationary point

    To find the nature of these stationary points, we find f”(x)

    f”(x) = 36x2 – 24x – 24

    When x = 0, f”(0)  =  36(0)2 – 24(0) – 24  =  -24 < 0

    Hence the curve will concave downwards, and (0, 1) is a maximum turning point.

    When x = -1, f”(-1)  =  36(-1)2 – 24(-1) – 24

    =  36 + 24 – 24  =  36 > 0

    Hence the curve will concave upwards, and (-1, 36) is a minimum turning point.

    When x = 2, f”(2)  =  36(2)2 – 24(2) – 24

    =  144 – 48 – 24  =  72 > 0

    Hence the curve will concave upwards, and (2, -31) is a minimum turning point.

    Final answer:

    • (0, 1) is a maximum turning point.
    • (-1, 36) is a minimum turning point.
    • (2, -31) is a minimum turning point.

    stationary point for fourth degree function

  2. Consider the curve y = f(x) = 2x3 – 4y’ = 6x2

    For stationary point, y’ = 0. Hence

    6x2  =  0    doubleright  x = 0

    When x = 0, y = 2(0)3 – 4  =  -4

    Hence (0, -4) is a stationary point.

    To find the type of stationary point, we find f”(x)

    f”(x) = 12x

    When x = 0, f”(x) = 0.

    Hence (0, -4) is a possible point of inflection.

    Now check for the concavity at (0, -4)

    x -1 0 1
    f”(x) f”(x) = 12(-1)

        = -12 < 0
    0 f”(x) = 12(1)

        = 12 > 0

    Since the concavity of the curve changes (0, -4) is a horizontal point of inflection.

    horizontal point of inflection

 

Here’s a summary table to help you sketch a curve using the first and second derivatives.

 

summary of stationary points