Stationary Points

(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: worksheet

Answers Without Working

  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: worksheet

Answers Without Working

  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
    \[\begin{pmatrix} 5,10\end{pmatrix}\]