The Second Derivative Test states:
Suppose the second derivative of a function is continuous on
(a, b) and c is in this interval.
- If df(c)dx = 0 and df2(c)/dx2 > 0, then the
function has a local minimum at c.
- If df(c)dx = 0 and df2(c)/dx2 < 0, then the
function has a local maximum at c.
The second derivative test can be better understood by continuing the
In the previous example,
the critical points is found when the derivative is 0, and this satisfies
the condition of the second derivative test, df(c)dx = 0. In
other words, the value of the first derivative of the function at 7 and
-1 is 0. In order to use second
at x = 7
and x = -1 are needed. Take another derivative to the first derivative
equation, df(x)/dx = 3x2 -
18x - 24, to obtain the
second derivative. That is
df2(x)/dx2 = df(df(x)/dx))/dx
= df(3x2 - 18x - 24)/dx
d(3x2)/dx - d(18x)/dx - d(24)/dx
= 6x -18
When x = 7
df2(x)/dx2 = 6x
-18 = 24 > 0
So the function has a local minimum at this critical point.
When x = -1
df2(x)/dx2 = 6x -18
= -24 < 0
So the function has a local maximum at this critical point. This result
is confirmed by the Function f(x) = x3 -
9x2 - 24x Diagram.