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Contents
Second Derivative Test
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Second
Derivative Test
The
second method of determining the nature of stationary points is by using the
second derivative.
Given
the point (a, f(a)) in the equation y = f(x) is a stationary point,
Maximum turning point
This
is when d2y/dx2
|x=a < 0
Why?
In the neighbourhood of the maximum turning point, the rate of change of
gradient is decreasing. Hence d2y/dx2
is
negative.
Minimum turning point
This
is when d2y/dx2
|x=a > 0
Conversely,
in the neighbourhood of the minimum turning point, the rate of change of
gradient is increasing. Hence d2y/dx2 is positive.
If
the second derivative test yields zero, the result is inconclusive. It does
not necessarily mean a stationary point of inflexion. In this case, the
first derivative test is used. The first derivative test should also be used
if it is tedious to find the second derivative.
Counter-example:
 Determine
the nature of the stationary point of the graph y = x4.
Solution:
dy/dx = 4x3
At stationary point, dy/dx = 0
4x3 = 0
x = 0
Second
derivative test:
d2y/dx2 = 12x2
d2y/dx2 |x=0 = 0
Is
it a point of inflexion? Or is it something else? We'll see with the first
derivative test.
First
derivative test:
For x = 0-, dy/dx < 0
For x = 0+, dy/dx > 0
The point is a minimum point.
Hence,
when the second derivative test yields zero, the point may not be a point of
inflexion.
Example:
Determine
the nature of the stationary points in the equation y = e2x (x - 3)2.
Solution:
dy/dx = 2e2x (x - 3)2 + 2(x - 3) e2x
= 2e2x (x - 3)(x - 2)
At stationary points, dy/dx = 0
2e2x (x - 3)(x - 2) = 0
x = 2 or x = 3
d2y/dx2 = 4e2x (x - 3)(x - 2) + 2e2x
(x - 3) + 2e2x (x - 2)
= 2e2x [2(x - 3)(x - 2) + (x - 3) + (x - 2)]
= 2e2x (2x2 - 8x + 7)
When
x = 2, d2y/dx2 |x=2 < 0
The
point is a maximum turning point.
When
x = 3, d2y/dx2 |x=3 > 0
The
point is a minimum turning point.
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