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

    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.

   

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

   For stationary point, y’ = 0. Hence

   6x²  =  0      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.

   

 

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