After establishing the
critical points of a function, the
second-derivative test uses the value of the
second derivative at those points to determine whether such points are a local
maximum or a local minimum. If the function
f is twice-differentiable at a critical point
x (i.e. a point where ''
(x'') = 0), then: • If f''(x) , then f has a local maximum at x. • If f''(x) > 0, then f has a local minimum at x. • If f''(x) = 0, the test is inconclusive. In the last case,
Taylor's theorem may sometimes be used to determine the behavior of
f near
x using
higher derivatives.
Proof of the second-derivative test Suppose we have f
(x) > 0 (the proof for f(x) is analogous). By assumption, f'(x) = 0. Then : 0 Thus, for
h sufficiently small we get : \frac{f'(x + h)}{h} > 0, which means that f'(x + h) if h (intuitively,
f is decreasing as it approaches x from the left), and that f'(x + h) > 0 if h > 0 (intuitively,
f is increasing as we go right from
x). Now, by the
first-derivative test, f has a local minimum at x.
Concavity test A related but distinct use of second derivatives is to determine whether a function is
concave up or concave down at a point. It does not, however, provide information about
inflection points. Specifically, a twice-differentiable function
f is concave up if f
(x) > 0 and concave down if f(x) . Note that if f(x) = x^4, then x = 0 has zero second derivative, yet is not an inflection point, so the second derivative alone does not give enough information to determine whether a given point is an inflection point.
Higher-order derivative test The
higher-order derivative test or
general derivative test is able to determine whether a function's critical points are maxima, minima, or points of inflection for a wider variety of functions than the second-order derivative test. As shown below, the second-derivative test is mathematically identical to the special case of
n = 1 in the higher-order derivative test. Let
f be a real-valued, sufficiently differentiable function on an interval I \subseteq \R, let c \in I, and let n \ge 1 be a
natural number. Also let all the derivatives of
f at
c be zero up to and including the
n-th derivative, but with the (
n + 1)th derivative being non-zero: : f'(c) = \cdots =f^{(n)}(c) = 0\quad \text{and}\quad f^{(n+1)}(c) \ne 0. There are four possibilities, the first two cases where
c is an extremum, the second two where
c is a (local) saddle point: • If
(n+1) is
even and f^{(n+1)}(c) , then
c is a local maximum. • If
(n+1) is even and f^{(n+1)}(c) > 0, then
c is a local minimum. • If
(n+1) is
odd and f^{(n+1)}(c) , then
c is a strictly decreasing point of inflection. • If
(n+1) is odd and f^{(n+1)}(c) > 0, then
c is a strictly increasing point of inflection. Since
(n+1) must be either odd or even, this analytical test classifies any stationary point of
f, so long as a nonzero derivative shows up eventually, where f^{(n+1)}(c) \ne 0. is the first non-zero derivative.
Example Say we want to perform the general derivative test on the function f(x) = x^6 + 5 at the point x = 0. To do this, we calculate the derivatives of the function and then evaluate them at the point of interest until the result is nonzero. : f'(x) = 6x^5, f'(0) = 0; : f
(x) = 30x^4, f(0) = 0; : f^{(3)}(x) = 120x^3, f^{(3)}(0) = 0; : f^{(4)}(x) = 360x^2, f^{(4)}(0) = 0; : f^{(5)}(x) = 720x, f^{(5)}(0) = 0; : f^{(6)}(x) = 720, f^{(6)}(0) = 720. As shown above, at the point x = 0, the function x^6 + 5 has all of its derivatives at 0 equal to 0, except for the 6th derivative, which is positive. Thus
n = 5, and by the test, there is a local minimum at 0. ==Multivariable case==