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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.

Tutorial - Introduction to Stationary Points

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

Answers Without Working

  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 \(\begin{pmatrix}1,0\end{pmatrix}\).

Solution With Working

Exercise 2

Solution Without Working

Solution With Working