Using the Second Derivative
to classify critical points as
maximums or minimums.
Let y = f(x) be a function. If the second derivative f''(x) is non-zero at a critical point x = a (that is a point where the first derivative is zero - f'(a) = 0) then the sign of the second derivative may be used to classify the point x = a as a local maximum or a local minimum.
Second Derivative Test:
Suppose y = f(x) and f'(a) = 0 and the second derivative f"(x) is continuous about an interval centered at x = a. Then
- x = a is a locally a minimum if f"(a) > 0,
- x = a is a locally a maximum if f"(a) < 0; and
Here if f"(a) = 0, the sign of f"(x) gives no result.
Memory Aid for what the sign of f"(a) implies: For f(x) = x2 has f"(x) = 2 > 0 and the critical point x = 0 is a minimum.
Example 1: Classification of Critical Points as Max or Min from sign of second derivative.Remark: When there are many maxima, the first and second derivative based tests alone can not say which the greatest or absolute maximum. For that function values are required. When there are many minima, the first and second derivative based tests alone can not say which the greatest or absolute minimum. For that function values are required.
"Proof" of the 2nd Derivative Test:
- If f'(a) = 0 and f"(a) > 0 then linear approximation of y' =f '(x) by f''(a)(x-a) implies near x = a that y' = f '(x) is negative for x < a and positive for x < a, and hence x = a is local minimum due to the first derivative test.
- If f'(a) = 0 and f"(a) < 0 then linear approximation of y' =f '(x) by f''(a)(x-a) implies near x = a that y' = f '(x) is positive for x < a and negative for x < a, and hence x = a is local maximum because of the first derivative test.
See your textbook (or the linear approximation theorem in site pages) for a fuller explanation of the second derivative test.
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