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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 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.

• (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. 