# Stationary Points (Critical Points)

Given a function $$f(x)$$ and its curve $$y=f(x)$$, a stationary point (aka critical point) is a point along the curve at which: $f'(x)=0$ Which can also be written: $\frac{dy}{dx} = 0$ Consequently at a stationary point the curve's gradient is equal to zero and the tangent to the curve horizontal.

A stationary point can be either one of the following three types:

• Maximum point (local or global)
• Minimum point (local or global)
• Horizontal Point of Inflexion
Stationary points are further introduced and illustrated in the following tutorial. Watch it now.

## Finding Stationary Points

Now that we have defined what a stationary point is we learn how to locate them.
We start by summarizing the method, below, in two steps. We then illustrate this method with a tutorial.

### Method - Finding Stationary Points

To find the stationary point(s) of a curve, $$y=f(x)$$, we need to determine the point(s) at which $$f'(x)=0$$. We can summarize the method in two steps:

• Step 1: find $$f'(x)$$.

• Step 2: solve $$f'(x)=0$$. The solution(s) $$x_1$$, $$x_2$$, ... to $$f(x)=0$$ are the $$x$$-coordinates of the stationary points of the curve.
At times we'll also have to calculate the $$y$$-coordinate(s) of the stationary point(s) in which case we would calculate them using the function $$f(x)$$. That's $$y_1=f(x_1)$$, $$y_2 = f(x_2)$$ ... .

This is further explained and illustrated in the following tutorial.

## Tutorial - Finding Stationary Points

In the following tutorial we review the ... .

## Exercise 1

Using the method, described above, find the coordinates of any stationary points on each of the following curves:

1. $$y = 2x^2 - 8x + 7$$

2. $$y = 2x^3 - 15x^2 + 2x + 1$$

3. $$y=-x^3-3x^2+çx +1$$

4. $$y = x + \frac{1}{x}$$
5. $$y = \frac{x^2}{2} - ln(x)+4$$

1. One stationary point at $$\begin{pmatrix}2,-1\end{pmatrix}$$.

2. Two stationary points: $$\begin{pmatrix}1,-1 \end{pmatrix}$$ and $$\begin{pmatrix}4,3\end{pmatrix}$$.

3. Two stationary points: $$\begin{pmatrix}-1,0\end{pmatrix}$$ and $$\begin{pmatrix}1,0\end{pmatrix}$$.