Derivative test

Stated precisely, suppose that f is a real-valued function defined on some open interval containing the point x and suppose further that f is continuous at x.

• If there exists a positive number r > 0 such that f is weakly increasing on (x − r, x] and weakly decreasing on [x, x + r), then f has a local maximum at x.
• If there exists a positive number r > 0 such that f is strictly increasing on (x − r, x] and strictly increasing on [x, x + r), then f is strictly increasing on (x − r, x + r) and does not have a local maximum or minimum at x.

Note that in the first case, f is not required to be strictly increasing or strictly decreasing to the left or right of x, while in the last case, f is required to be strictly increasing or strictly decreasing. The reason is that in the definition of local maximum and minimum, the inequality is not required to be strict: e.g. every value of a constant function is considered both a local maximum and a local minimum.

The first-derivative test depends on the "increasing–decreasing test", which is itself ultimately a consequence of the mean value theorem. It is a direct consequence of the way the derivative is defined and its connection to decrease and increase of a function locally, combined with the previous section.

Suppose f is a real-valued function of a real variable defined on some interval containing the critical point a. Further suppose that f is continuous at a and differentiable on some open interval containing a, except possibly at a itself.

• If there exists a positive number r > 0 such that for every x in (a − r, a) we have f′(x) ≥ 0, and for every x in (a, a + r) we have f′(x) ≤ 0, then f has a local maximum at a.
• If there exists a positive number r > 0 such that for every x in (a − r, a) we have f′(x) ≤ 0, and for every x in (a, a + r) we have f′(x) ≥ 0, then f has a local minimum at a.
• If there exists a positive number r > 0 such that for every x in (a − r, a) ∪ (a, a + r) we have f′(x) > 0, then f is strictly increasing at a and has neither a local maximum nor a local minimum there.
• If none of the above conditions hold, then the test fails. (Such a condition is not vacuous; there are functions that satisfy none of the first three conditions, e.g. f(x) = x² sin(1/x)).

Again, corresponding to the comments in the section on monotonicity properties, note that in the first two cases, the inequality is not required to be strict, while in the third, strict inequality is required.

The first-derivative test is helpful in solving optimization problems in physics, economics, and engineering. In conjunction with the extreme value theorem, it can be used to find the absolute maximum and minimum of a real-valued function defined on a closed and bounded interval. In conjunction with other information such as concavity, inflection points, and asymptotes, it can be used to sketch the graph of a function.

Suppose we have ${\displaystyle f''(x)>0}$ (the proof for ${\displaystyle f''(x)<0}$ is analogous). By assumption, ${\displaystyle f'(x)=0}$. Then

${\displaystyle 0<f''(x)=\lim _{h\to 0}{\frac {f'(x+h)-f'(x)}{h}}=\lim _{h\to 0}{\frac {f'(x+h)}{h}}.}$

Thus, for h sufficiently small we get

${\displaystyle {\frac {f'(x+h)}{h}}>0,}$

which means that ${\displaystyle f'(x+h)<0}$ if ${\displaystyle h<0}$ (intuitively, f is decreasing as it approaches ${\displaystyle x}$ from the left), and that ${\displaystyle f'(x+h)>0}$ if ${\displaystyle h>0}$ (intuitively, f is increasing as we go right from x). Now, by the first-derivative test, ${\displaystyle f}$ has a local minimum at ${\displaystyle x}$.

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 ${\displaystyle f''(x)>0}$ and concave down if ${\displaystyle f''(x)<0}$. Note that if ${\displaystyle f(x)=x^{4}}$, then ${\displaystyle x=0}$ 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.

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 ${\displaystyle I\subset \mathbb {R} }$, let ${\displaystyle c\in I}$, and let ${\displaystyle n\geq 1}$ 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:

${\displaystyle f'(c)=\cdots =f^{(n)}(c)=0\quad {\text{and}}\quad f^{(n+1)}(c)\neq 0.}$

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 is odd and ${\displaystyle f^{(n+1)}(c)<0}$, then c is a local maximum.
• If n is odd and ${\displaystyle f^{(n+1)}(c)>0}$, then c is a local minimum.
• If n is even and ${\displaystyle f^{(n+1)}(c)<0}$, then c is a strictly decreasing point of inflection.
• If n is even and ${\displaystyle f^{(n+1)}(c)>0}$, then c is a strictly increasing point of inflection.

Since n must be either odd or even, this analytical test classifies any stationary point of f, so long as a nonzero derivative shows up eventually.

Say we want to perform the general derivative test on the function ${\displaystyle f(x)=x^{6}+5}$ at the point ${\displaystyle x=0}$. To do this, we calculate the derivatives of the function and then evaluate them at the point of interest until the result is nonzero.

${\displaystyle f'(x)=6x^{5}}$, ${\displaystyle f'(0)=0;}$

${\displaystyle f''(x)=30x^{4}}$, ${\displaystyle f''(0)=0;}$

${\displaystyle f^{(3)}(x)=120x^{3}}$, ${\displaystyle f^{(3)}(0)=0;}$

${\displaystyle f^{(4)}(x)=360x^{2}}$, ${\displaystyle f^{(4)}(0)=0;}$

${\displaystyle f^{(5)}(x)=720x}$, ${\displaystyle f^{(5)}(0)=0;}$

${\displaystyle f^{(6)}(x)=720}$, ${\displaystyle f^{(6)}(0)=720.}$

As shown above, at the point ${\displaystyle x=0}$, the function ${\displaystyle x^{6}+5}$ 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.