A stationary point is a point on a curve where the gradient is zero ($0$). This means that the derivative will be $0$: $$ \begin{equation}\begin{aligned} \frac{dy}{dx}=0\\ \end{aligned}\end{equation} $$
To find the stationary points, we simply find $dy/dx$ and equate it to $0$. We then solve this equation to get the $x$ value(s) that satisfy it. The $y$ values can be found by substituting these $x$ values into the original equation of the curve.
Types of stationary points
There are three ($3$) types of stationary points:
- Local maximum points
- Local minimum points
- Horizontal points of inflection
Note that we say local maximum and minimum points because these turning points are higher or lower relative to the points that come immediately before and after them but are not necessarily the highest or lowest points on the curve. Such highest or lowest points are referred to as global maximum and minimum points.
Example 1
Find the stationary points on the curve $y=x^2+5x+6$.
Solution The derivative is $$ \begin{equation}\begin{aligned} y&=x^2+5x+6\\ \therefore \frac{dy}{dx}&=2x+5\\ \end{aligned}\end{equation} $$
We then equate this to $0$ and solve: $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=0\\ \therefore 2x+5&=0\\ 2x&=-5\\ x&=\frac{-5}{2}\\ \end{aligned}\end{equation} $$
The $y$ values can be found by using the original equation: $$ \begin{equation}\begin{aligned} y&=x^2+5x+6\\ y&=(\frac{-5}{2})^2+5(\frac{-5}{2})+6\\ y&=\frac{25}{4}+\frac{-25}{2}+6\\ y&=\frac{25-50+24}{4}\\ y&=\frac{-1}{4}\\ \end{aligned}\end{equation} $$
Thus the stationary point is $(\frac{-5}{2},\frac{-1}{4})$.
Example 2
Find the stationary points on the curve $y=x^3-12x$.
Solution The derivative is $$ \begin{equation}\begin{aligned} y&=x^3-12x\\ \therefore \frac{dy}{dx}&=3x^2-12\\ \end{aligned}\end{equation} $$
We then equate this to $0$ and solve: $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=0\\ \therefore 3x^2-12&=0\\ x^2-4&=0\\ (x+2)(x-2)&=0\\ x&=-2\text{ OR }x=2\\ \end{aligned}\end{equation} $$
For $x=2$ the $y$ value is $$ \begin{equation}\begin{aligned} y&=(2)^3-12(2)\\ y&=-16\\ \end{aligned}\end{equation} $$
For $x=-2$ the $y$ value is $$ \begin{equation}\begin{aligned} y&=(-2)^3-12(-2)\\ y&=16\\ \end{aligned}\end{equation} $$
Thus the stationary points on the curve are $(-2,16)$ and $(2,-16)$.
The second derivative test
This test is used to determine the nature of the turning point for maximum and minimum curves:
- When $\frac{d^2y}{dx^2}>0$, we have a minimum point
- When $\frac{d^2y}{dx^2}<0$, we have a maximum point
- When $\frac{d^2y}{dx^2}=0$, it is indeterminate
Notice how the value for $\frac{d^2y}{dx^2}$ behaves like the $a$ in $ax^2+bx+c$. Thus if we know what $a$ tells us about the nature of a turning point for a quadratic curve, we know the second derivative test.
Example
Determine the nature of the stationary points of the curve $y=x^3-12x$.
Solution From the derivative we find the second derivative: $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=3x^2-12\\ \therefore \frac{d^2y}{dx^2}&=6x\\ \end{aligned}\end{equation} $$
When $x=-2$: $$ \begin{equation}\begin{aligned} \frac{d^2y}{dx^2}&=6(-2)\\ &=-12\rightarrow \text{-ve}\\ \end{aligned}\end{equation} $$
Thus at the point $(-2,16)$, there is a maximum turning point. When $x=2$: $$ \begin{equation}\begin{aligned} \frac{d^2y}{dx^2}&=6(2)\\ &=12\rightarrow \text{+ve}\\ \end{aligned}\end{equation} $$
Thus at the point $(2,-16)$, there is a minimum turning point.