# Stationary Points - Part 1

## (Definition & How to Find Stationary Points)

A stationary point, or critical point, is a point at which the curve's gradient equals to zero. Consequently if a curve has equation $$y=f(x)$$ then at a stationary point we'll always have: $f'(x)=0$ which can also be written: $\frac{dy}{dx} = 0$ In other words the derivative function equals to zero at a stationary point.

## Different Types of Stationary Points

There are three types of stationary points:

• local (or global) maximum points
• local (or global) minimum points
• horizontal (increasing or decreasing) points of inflexion.
It is worth pointing out that maximum and minimum points are often called turning points.

### Turning Points

A turning point is a stationary point, which is either:

• a local (or global) minimum
• a local (or global) maximum
each of which are illustared in the graphs shown here, where the horizontal tangent is shown in orange:

### Horizontal Points of Inflection

A horizontal point of inflection is a stationary point, which is either:

• a increasing horizontal point of inflection
• a decreasing horizontal point of inflection
each of which are illustared in the graphs shown here, where the horizontal tangent is shown in orange:

## Method: finding stationary points

Given a function $$f(x)$$ and its curve $$y=f(x)$$, to find any stationary point(s) we follow three steps:

• Step 1: find $$f'(x)$$
• Step 2: solve the equation $$f'(x)=0$$, this will give us the $$x$$-coordinate(s) of any stationary point(s).
• Step 3 (if needed/asked): calculate the $$y$$-coordinate(s) of the stationary point(s) by plugging the $$x$$ values found in step 2 into $$f(x)$$.

## Tutorial 1

In the following tutorial we illustrate how to use our three-step method to find the coordinates of any stationary points along a function's curve's length.

## Example 1

Given the function defined by the equation: $y = x^2 - 4x+5$ find the coordinates of any stationary point(s).

### Solution

Following our three-step method:

• Step 1: find $$\frac{dy}{dx}$$.
For $$y = x^2 - 4x+5$$, we find: $\frac{dy}{dx} = 2x-4$
• Step 2: solve the equation $$\frac{dy}{dx}=0$$.
That's: $2x - 4 = 0$ Solving this leads to: $2x = 4$ Finally: $x = 2$ At this stage we can state the curve $$y=x^2 - 4x + 5$$ has one stationary point whose $$x$$-coordinate is $$x=2$$.
• Step 3: calculate the sationary point's $$y$$-coordinate.
To do this we replace $$x$$ by $$2$$ in the curve's equation $$y = x^2 - 4x + 5$$.
That's: \begin{aligned} y & = 2^2 - 4 \times 2 + 5 \\ & = 4 - 8 + 5 \\ & = -4 + 5 \\ y & = 1 \end{aligned} So the stationary point has $$y$$-coordinate $$y=1$$.
We can therefore state that the curve $$y = x^2 - 4x + 5$$ has one stationary point with coordinates $$\begin{pmatrix}2,1 \end{pmatrix}$$.

This result is confirmed, using our graphical calculator and looking at the curve $$y=x^2 - 4x+5$$:

We can see quite clearly that the curve has a global minimum point, which is a stationary point, at $$\begin{pmatrix}2,1 \end{pmatrix}$$.

## Example 2

Find the coordinates of any stationary point(s) of the function defined by: $y = 2x^3 + 3x^2 - 12x+1$

### Solution

Following our three-step method:

• Step 1: find $$\frac{dy}{dx}$$.
Using the power rule for differentiation we find: \begin{aligned} \frac{dy}{dx} & = 3\times 2x^{3-1} + 2\times 3x^{2-1} - 12 \\ & = 6x^2+6x^1 - 12 \\ \frac{dy}{dx} & = 6x^2+6x - 12 \end{aligned}
• Step 2: solve the equation $$\frac{dy}{dx}=0$$.
Since $$\frac{dy}{dx} =6x^2+6x - 12$$ we need to solve the quadratic equation: $6x^2+6x-12 = 0$ Doing this either using our graphical calculator, the quadratic formula, or by factoring, we find two solutions: $x = -2 \quad \text{and} \quad x = 1$ So at this stage we can state that the function $$y = 2x^3+3x^2-12x+1$$ has two stationary points. One with $$x$$-coordinate $$x=-2$$ and the other with $$x$$-coordinate $$x=1$$.
• Step 3: calculate the sationary point's $$y$$-coordinate.
Since we found two values of $$x$$, in step 2, there are two $$y$$-coordinates to calculate, one for each value of $$x$$.
• when $$x=-2$$:
Replacing $$x$$ by $$-2$$ in $$y = 2x^3+3x^2-12x+1$$, we find: \begin{aligned} y & = 2\times (-2)^3+3\times (-2)^2 - 12 \times (-2) + 1 \\ & = 2\times (-8) + 3\times 4 - (-24)+1\\ & = -16 + 12 + 24 + 1 \\ y & = 21 \end{aligned} So the function has a stationary point at: $\begin{pmatrix}-2,21 \end{pmatrix}$
• when $$x=1$$:
Replacing $$x$$ by $$1$$ in $$y = 2x^3+3x^2-12x+1$$, we find: \begin{aligned} y & = 2\times 1^3+3\times 1^2 - 12 \times 1 + 1 \\ & = 2\times 1 + 3\times 1 - 12+1\\ & = 2+3-12+1 \\ y & = -6 \end{aligned} So the function has its second stationary point at: $\begin{pmatrix}1,-6 \end{pmatrix}$
We can see both of these stationary points on the graph shown below:

We can see quite clearly that the stationary point at $$\begin{pmatrix}-2,21\end{pmatrix}$$ is a local maximum and the stationary point at $$\begin{pmatrix}1,-6\end{pmatrix}$$ is a local minimum.

## Example 3

Given the function defined by: $y = x^3-6x^2+12x-12$ Find the coordinates of any stationary point(s) along this function's curve's length.

### Solution

Following our three-step method:

• Step 1: find $$\frac{dy}{dx}$$.
Since $$y = x^3-6x^2+12x-12$$, we use the power rule for differentiation, to find this function's derivative: \begin{aligned} \frac{dy}{dx} & = 3\times x^{3-1} - 2 \times 6x^{2-1} + 12x^{1-1} + 0 \\ & = 3x^2 - 12x^1 + 12x^0 \\ \frac{dy}{dx} & = 3x^2 - 12x + 12 \end{aligned}
• Step 2: solve the equation $$\frac{dy}{dx}=0$$.
Since $$\frac{dy}{dx} = 3x^2 - 12x + 12$$ we have to solve the quadratic equation: $3x^2 - 12x + 12= 0$ We can solve this using the quadratic formula or by factoring or even using our graphical calculator.
In doing so we find one solution: $x = 2$ So, at this stage, we can state that this function has one stationary point whose $$x$$-coordinate is $$x = 2$$.
• Step 3: calculate the sationary point's $$y$$-coordinate.
Since we found that the stationary point had $$x$$-coordinate $$x = 2$$, to find its $$y$$-coordinate we replace $$x$$ by $$2$$ in the function's equation $$y = x^3-6x^2+12x-12$$. That's: \begin{aligned} y & = 2^3 - 6\times 2^2 + 12 \times 2 - 12 \\ & = 8 - 6\times 4 + 24 - 12 \\ & = 8 - 24 + 24 - 12 \\ y& = -4 \end{aligned} So this function has a stationary point with coordinates: $\begin{pmatrix}2,-4\end{pmatrix}$
This result is confirmed when we look at the graph of $$y = x^3 - 6x^2 + 12x - 12$$:

Looking at this graph, we can see that this curve's stationary point at $$\begin{pmatrix}2,-4\end{pmatrix}$$ is an increasing horizontal point of inflection.

## Exercise 1

Find the coordinates of any stationary point(s) along the length of each of the following curves:

1. $$y = x^2 - 2x - 8$$
2. $$y = -x^2-6x-8$$
3. $$y = 2x^3 - 12x^2 - 30x - 10$$
4. $$y = -2x^3 + 3x^2 + 36x - 6$$
5. $$y = x^3+3x^2+3x-2$$

Note: this exercise can be downloaded as a worksheet to practice with:

1. We find the derivative to be $$\frac{dy}{dx} = 2x-2$$ and this curve has one stationary point: $\begin{pmatrix} 1,-9\end{pmatrix}$
2. We find the derivative to be $$\frac{dy}{dx} = -2x-6$$ and this curve has one stationary point: $\begin{pmatrix} -3,1\end{pmatrix}$
3. We find the derivative to be $$\frac{dy}{dx} = 2x^3 - 12x^2 - 30x- 10$$ and this curve has two stationary points: $\begin{pmatrix} -1,6\end{pmatrix}$
and
$\begin{pmatrix} 5,-210\end{pmatrix}$
4. We find the derivative to be $$\frac{dy}{dx} = -2x^3+3x^2+36x - 6$$ and this curve has two stationary points: $\begin{pmatrix} -2,-50\end{pmatrix}$
and
$\begin{pmatrix} 3,75\end{pmatrix}$
5. We find the derivative to be $$\frac{dy}{dx} = x^3+3x^2+3x-2$$ and this curve has one stationary point: $\begin{pmatrix} -1,-3\end{pmatrix}$

## Tutorial 2

In the following tutorial we illustrate how to use our three-step method to find the coordinates of any stationary points along a function's curve's length for a function whose type common in exams: $$y = ax + \frac{b}{x}$$.

## Example 4

Given the function defined by: $y = x+\frac{4}{x}$ find the coordinates of any stationary points along this curve's length.

### Solution

Following our three-step method:

• Step 1: find $$\frac{dy}{dx}$$.
We can re-write $$y=x+\frac{4}{x}$$, using negative exponents: $y = x+4.x^{-1}$ We can now use the power rule for differentiation to find the derivative: \begin{aligned} \frac{dy}{dx} & = 1+(-1).4.x^{-1-1} \\ & = 1-4x^{-2} \\ \frac{dy}{dx} & = 1 - \frac{4}{x^2} \end{aligned}
• Step 2: solve the equation $$\frac{dy}{dx}=0$$.
Since $$\frac{dy}{dx} =1 - \frac{4}{x^2}$$ we need to solve the equation: $1 - \frac{4}{x^2} = 0$ To solve this equation by hand we start by writing the entire left hand side over $$x^2$$ using the fact that $$1 = \frac{x^2}{x^2}$$ so that: $1 - \frac{4}{x^2} = 0$ can be written: $\frac{x^2}{x^2} - \frac{4}{x^2} = 0$ and therefore we have to solve: $\frac{x^2-4}{x^2} = 0$ This will equal to $$0$$ if and only if the numerator, $$x^2-4$$, equals to $$0$$.
So all we need to solve is: $x^2 - 4 = 0$ That's: $x^2 = 4$ which leads to two solutions: $x = -2 \quad \text{and} \quad x = 2$
• Step 3: calculate the sationary point's $$y$$-coordinate.
Since we found two values of $$x$$, in step 2, there are two $$y$$-coordinates to calculate, one for each value of $$x$$.
• when $$x = -2$$:
replacing $$x$$ by $$-2$$ in $$y = x+\frac{4}{x}$$, we find: \begin{aligned} y & = -2 + \frac{4}{-2} \\ & = -2 + (-2) \\ & = -2 - 2 \\ y & = -4 \end{aligned} So one of this function's stationary points is: $\begin{pmatrix}-2,-4 \end{pmatrix}$
• when $$x = 2$$:
replacing $$x$$ by $$2$$ in $$y = x+\frac{4}{x}$$, we find: \begin{aligned} y & = 2 + \frac{4}{2} \\ & = 2 + 2 \\ y & = 4 \end{aligned} So the function's second stationary point has coordinates: $\begin{pmatrix}2,4 \end{pmatrix}$
We can see both of these stationary points on the graph shown below:

We can see quite clearly that the stationary point at $$\begin{pmatrix}-2,-4\end{pmatrix}$$ is a local maximum and the stationary point at $$\begin{pmatrix}2,4\end{pmatrix}$$ is a local minimum.

## Exercise 2

Find the coordinates of any stationary point(s) along the length of each of the following curves:

1. $$y = 2x + \frac{8}{x}$$
2. $$y = -x - \frac{1}{x}$$
3. $$y = 3x + \frac{27}{x}$$
4. $$y = -2x - \frac{72}{x}$$
5. $$y = x + \frac{25}{x}$$

Note: this exercise can be downloaded as a worksheet to practice with:

1. We find the derivative to be $$\frac{dy}{dx} = 2 - \frac{8}{x^2}$$ and this curve has two stationary points: $\begin{pmatrix} -2,-8\end{pmatrix}$
and
$\begin{pmatrix} 2,8\end{pmatrix}$
2. We find the derivative to be $$\frac{dy}{dx} = -1 + \frac{1}{x^2}$$ and this curve has two stationary points: $\begin{pmatrix} -1,2\end{pmatrix}$
and
$\begin{pmatrix} 1,-2\end{pmatrix}$
3. We find the derivative to be $$\frac{dy}{dx} = 3 - \frac{27}{x^2}$$ and this curve has two stationary points: $\begin{pmatrix} -3,-18\end{pmatrix}$
and
$\begin{pmatrix} 3,18\end{pmatrix}$
4. We find the derivative to be $$\frac{dy}{dx} = -22 + \frac{72}{x^2}$$ and this curve has two stationary points: $\begin{pmatrix} -6,48\end{pmatrix}$
and
$\begin{pmatrix} 6,-48\end{pmatrix}$
5. We find the derivative to be $$\frac{dy}{dx} = 1 - \frac{25}{x^2}$$ and this curve has two stationary points: $\begin{pmatrix} -5,-10\end{pmatrix}$
and
$\begin{pmatrix} 5,10\end{pmatrix}$