Stationary points

Stationary points are defined as:

For example, the stationary points of $f(x)=x^{2}-x-12$ occur at $f^{\prime}(x)=2 x-1=0$. Solving for $x$ gives the only stationary point at $x=1 / 2$. As another example consider $f(x)=1-\cos x$ with stationary points at $f^{\prime}(x)=\sin x=0$; this gives $x=n \pi$ where $n \in \mathbb{Z}$. Of course there are functions with no stationary points where the gradient is never zero like for instance $f(x)=x /(x+1)$ whose derivative is $f^{\prime}(x)=(x+1)^{-2}$.

Second derivative test

Here we discuss local and global minimum and maximum values of a function; these occur at stationary points. We start with the definitions of local extrema.

Note that if $f^{\prime}(a)=0$ we do not necessarily have a minimum or maximum point at $x=a$. At a minimum value, the function changes from decreasing to increasing while at a maximum, the function changes from increasing to decreasing. If the function has a zero slope at $x=a$ but it is either always increasing or always decreasing on either side of the point, we refer to it as a point of inflection. Coming back to the second derivative test:

if $f^{\prime \prime}(a)=0$, the test is inconclusive: $x=a$ may be a local minimum, a local maximum, or a point of inflection.

If $f^{\prime \prime}(a)=0$ and $f^{\prime \prime \prime}(a) \neq 0$ then $x=a$ is a point of inflection. Note that the converse is not true, i.e. if $x=a$ is a point of inflection then it does not follow that $f^{\prime \prime}(a)=0$ and $f^{\prime \prime \prime}(a) \neq 0$. For instance, $x=0$ is a point of inflection of $f(x)=x^{5}$ but $f^{\prime \prime \prime}(0)=0$.- Also, it is possible that $f^{\prime \prime}(a)=0$ at a point $a$ which is not a point of inflection, e.g. $f(x)=x^{4}$ at $x=0$.

Higher-order (or general) derivative test

Since the second derivative test is inconclusive for a stationary point $x=a$ where $f^{\prime \prime}(a)=0$, we need to use another test. The higher-order derivative test (also referred to as the general derivative test) can help us identify whether a function's critical points are maxima, minima, or inflection points. Note the slight subtlety in the definitions of critical and stationary points. All stationary points are critical points but the reverse is not true. A critical point $x=a$ occurs if $f(a)$ exists and wherever either $f^{\prime}(a)=0$ or $f^{\prime}(a)$ does not exist. If $f^{\prime}(a)$ does not exist then, it follows that $f$ is not differentiable at $x=a$ and $x=a$ is a singular point (as mentioned in the discussion in the differentiation section). Further, if $f^{\prime}(a)=0$, then the critical point is a stationary point (as stated in the definition of stationary points).

Consider a function $f(x)$ with the point $x=a$ in its domain and where $x=a$ is a stationary point. Suppose that the first $n$ derivatives of $f$ at $x=a$ are zero and the $n+1$ derivative of $f$ at $a$ exists and it is nonzero, i.e.

So, $f^{(n+1)}(a)$ is the first derivative that is nonzero-valued at the point $x=a$. Then, we have the following possibilities:

(i) If $n$ is odd and $f^{(n+1)}(a)<0$ : $a$ is a local maximum

(ii) If $n$ is odd and $f^{(n+1)}(a)>0: a$ is a local minimum

Note that the first two cases are equivalent to the second derivative test when $n=1$.

(iii) If $n$ is even and $f^{(n+1)}(a)<0: a$ is a point of inflection and the function is decreasing there.

(iv) If $n$ is even and $f^{(n+1)}(a)>0: a$ is a point of inflection and the function is increasing there.

When several derivatives of a function vanish at a point, the function becomes pretty flat there. Note that the second derivative test is mathematically equivalent to the special case of $n=1$ in the higher-order derivative test.

For example, say we wanted to determine the stationary points and inflection points of the graph of $y=x^{4}-2 x^{2}$. Stationary points occur at $f^{\prime}(x)=0$, so:

giving $x=0$ and $x= \pm 1$ as stationary points. The second derivative is $y^{\prime \prime}(x)=12 x^{2}-4$ which, when evaluated at each of the stationary points gives

according to the second derivative test, $x=0$ is a local maximum and $x= \pm 1$ are local minima. Further, $y^{\prime \prime}(x)=0$ at $x= \pm 1 / \sqrt{3}$ and $y^{\prime \prime \prime}( \pm 1 / \sqrt{3}) \neq 0$. Since $y^{\prime \prime}( \pm 1 / \sqrt{3})=0$ and $y^{\prime \prime \prime}( \pm 1 / \sqrt{3}) \neq 0$, the points $x \pm 1 / \sqrt{3}$ are points of inflection. Note however that since $y^{\prime}( \pm 1 / \sqrt{3}) \neq 0$, these are not stationary points.

Global extrema

So far we have talked about local minima and maxima. We require that a point $a$ is a local maximum (minimum) relative to some open interval of $x$ 's around $x=a$. That is to say that there may be larger (smaller) values somewhere else in the domain but local (i.e. very close) to $x=a, f(a)$ is larger (smaller) than all other values that $f$ can take.

At a global maximum (minimum), $f(a)$ is the largest (smallest) value $f$ can take. We illustrate the different types of extrema using the graph of a function $y=f(x)$:

The function has 3 stationary points at $x=-4,-1$, and 1 . The point at $\left(-4, y_{1}\right)$ is a local minimum since $f(-4)$ is smaller than $f$ at some $x$ in the neighbourhood of $x=-4$ (i.e. very close to $x=-4$, on both sides). Additionally, the point $\left(-1, y_{2}\right)$ is also a local minimum since $f(1)$ is smaller than $f$ at some $x$ in the neighbourhood of $x=1$. Since $f(-4)$ is the smallest value $f$ can take in the domain in which $f$ is defined, we also refer to $\left(-4, y_{1}\right)$ as a global minimum. There exists only one maximum at $\left(-1, y_{3}\right)$ and it is both a local and global maximum.