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:
- If f'(x) = 0 and f”(x) > 0, then there is a minimum turning point
- If f'(x) = 0 and f”(x) < 0, then there is a maximum turning point
- 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.
- 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.
- 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 < 00 f”(x) = 12(1) = 12 > 0Since 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.