Stationary points

A stationary point (also known as a critical point) is a point on the graph of a function where its derivative is zero or undefined. These points are important because they often correspond to local maxima, local minima, or points of inflection.  Understanding stationary points is key to analyzing and interpreting the behavior of functions.

Use this page to revise the following concepts within stationary points:

The nature of stationary points

To classify the nature of a stationary point, we have two main tests, the first derivative test and the second derivative test.

In the first derivative test, we calculate the derivative at points just to the left and right of each stationary point. The sign on the detivative will tell us the nature of the stationary point, as detailed in the following image.

A diagram illustrating different types of stationary points on a function, as defined by the results of the first derivative test. It includes four cases: a minimum with a negative slope (negative derivative) before and a positive slope (positive derivative) after, a maximum with a positive slope before and a negative slope after, a stationary point of inflection with negative slopes on either side, and a stationary point of inflection with positive slopes on either side

Alternatively, if the second derivative exists we may use second derivative test.  In this test we find the second derivative at the stationary point. The sign of the second derivative will tell us the nature of the stationary point.

A visual comparison of parabolas with minimum or maximum stationary points. On the left, a blue upward-opening parabola with the label "Concave up" has a minimum stationary point, and the statement "f''(x) is positive" is written below it. On the right, a green downward-opening parabola with the label "Concave down" has a maximum stationary point, and the statement "f''(x) is negative" is written below it. The image illustrates how the sign of the second derivative corresponds to the concavity of a function, and whether a stationary point is a maximum or minimum.

If the second derivative is equal to zero, then the stationary point is a point of inflection.

Local maximum

At a local maximum, the slope of \(f(x)\) is positive to the left of the stationary point and negative to the right of the stationary point. This is also referred to as a turning point.

A local maximum occurs if either of the following conditions are met:

First Derivative Test:

\(f^\prime\left(x\right) = 0\) and \(f^\prime\left(x - h\right) > 0\) and \(f^\prime\left(x + h\right) < 0\),   where \(h\) is very small.

or

Second Derivative Test:

\(f^\prime\left(x\right) = 0\) and \(f^{\prime\prime}\left(x\right) < 0\)

Note: \(f^{\prime\prime}(x)\) refers to the second derivative of \(f(x)\) – that is, the derivative of the derivative of \(f(x)\).

Worked Example

Find the local maximum for the function \(y=x^3-25x+10\)

\[\begin{align}f\left(x\right) &= x^{3} - 25x + 10 \\ f^\prime\left(x\right) &= 3x^{2} - 25 \\ f^\prime\left(x\right) &= 0 \\3x^{2} &= 25 \\ x^{2} &= \frac{25}{3} \\ x &= \pm\frac{5}{\sqrt{3}}\end{align}\]

A graph of a circular function f(x)=sine(x) (blue curve) alongside its derivative f'(x)=cos(x) (green curve). The function has a sinusoidal shape, with peaks and troughs. The derivative is also sinusoidal but phase-shifted, crossing zero at the maxima and minima of f(x). This illustrates how the derivative represents the slope of f(x), being positive on increasing intervals, negative on decreasing intervals, and zero at local maxima and minima.

By calculating \(f\left(\pm\frac{5}{\sqrt{3}}\right)\), we can see the turning points exist at

\[\begin{array}{II}\left(-\frac{5}{\sqrt{3}}, \frac{250}{3\sqrt{3}}+10\right)\ \textsf{or}\ \left(-2.89, 58.1\right) \\ \left(\frac{5}{\sqrt{3}}, -\frac{250}{3\sqrt{3}} + 10\right)\ \textsf{or}\ \left(2.89, -3.88.1\right)\end{array}\]

We can determine the point at \(\left(-\dfrac{5}{\sqrt{3}}, \dfrac{250}{3\sqrt{3}} + 10\right)\ \textsf{or}\ \left(-2.89, 58.2\right)\) is a local maximum by performing the first derivative test or second derivative test:

First derivative test:

\(f^′\left(\frac{−5}{\sqrt{3}}−1\right)=20.3\)
\(f^′\left(\frac{−5}{\sqrt{3}}\right)=0\)
\(f^′\left(\frac{−5}{\sqrt{3}}+1\right)=−14.3\)

A diagram illustrating a point is a local maximum if the slope and therefore the first derivative is positive on the left of the stationary point and the slope and the first derivative is negative on the right of the stationary point.

Second derivative test:

\(f^{\prime\prime}\left(x\right)=6x\)
\(f^{\prime\prime}\left(\frac{−5}{\sqrt{3}}−1\right)=-1.7\)		A diagram illustrating a point is a local maximum the second derivative is negative, meaning the parabola’s shape will be concave down.

Local minimum

At a local minimum, the slope of \(f(x)\) is negative to the left of the stationary point and positive to the right of the stationary point.

A local minimum occurs if either of the following conditions are met:

First Derivative Test:

\(f^\prime\left(x\right) = 0\) and \(f^\prime\left(x - h\right) > 0\) and \(f^\prime\left(x + h\right) < 0\), where \(h\) is very small.

or

Second Derivative Test:

\(f^\prime\left(x\right) = 0\) and \(f^{\prime\prime} > 0\)

Worked Example

\[\begin{align}f\left(x\right) &= x^{3} - 25x + 10 \\ f^\prime\left(x\right) &= 3x^{2} - 25 \\ f^\prime\left(x\right) &= 0\\ 3x^{2} &= 25  \\ x^{2} &= \frac{25}{3}\\ x &= \pm\frac{5}{\sqrt{3}}\end{align}\]

A graph of the cubic function f(x) = x^3 - 25x + 10. The graph exhibits typical cubic behaviour, with the function increasing, decreasing, then increasing again. The curve has three critical points, a local maximum at (-2.89, 58.1) and a local minimum at (2.89, 38.1), where the slope of the function is zero, and a point of inflection at (0, 10).

The turning points are

\[\begin{array}{II}\left(-\frac{5}{\sqrt{3}}, \frac{250}{3\sqrt{3}} + 10\right)\ \textsf{or}\ \left(-2.89, 58.1\right)  \\ \left(\frac{5}{\sqrt{3}}, -\frac{250}{3\sqrt{3}} + 10\right)\ \textsf{or}\ \left(2.89, -38.1\right)\end{array}\]

We can determine the point at  \(\left(\dfrac{5}{\sqrt{3}}, -\dfrac{250}{3\sqrt{3}}+10\right)\ \textsf{or}\ \left(2.89, -38.1\right)\) is a local minimum by performing the first derivative test or second derivative test:

First derivative test:

\(f^′\left(-\frac{−5}{\sqrt{3}}−1\right)=−14.3\)
\(f^′\left(-\frac{−5}{\sqrt{3}}\right)=0\)
\(f^′\left(-\frac{−5}{\sqrt{3}}+1\right)=20.3\)

A diagram illustrating a point is a local minimum if the slope and therefore the first derivative is negative on the left of the stationary point and the slope and the first derivative is positive on the right of the stationary point.

Second derivative test:

\(f^{\prime\prime}\left(x\right)=6x\)
\(f^{\prime\prime}\left(-\frac{−5}{\sqrt{3}}−1\right)=1.7\)		A diagram illustrating a point is a local maximum the second derivative is negative, meaning the parabola’s shape will be concave down.

Stationary point of inflection

At a stationary point of inflection , the slope of \(f(x)\) is zero, and the curve changes concavity on either side of this point , but the slope continues to be of the same sign.

A stationary point of inflection occurs if any of the following conditions are met:

First Derivative Test:

\(f^\prime\left(x\right) = 0\) and \(f^\prime\left(x - h\right) > 0\) and \(f^\prime\left(x + h\right) > 0\) where \(h\) is very small.

That is, there is no sign change in \(f'(x)\) across the point.

or

Second Derivative Test:

\(f^\prime\left(x\right) = 0\) and \(f^{\prime\prime}\left(x\right) = 0\)

And, there is a sign change in \(f^{\prime\prime}\left(x\right)\) across the point

Worked Example

Example 1

Determine the stationary point of inflection in \(f\left(x\right) = x^{3} - 15x^{2} + 75x - 525\).

\(f^\prime\left(x\right) = 3x^{2} - 30x +75\)

Solve \(f^\prime\left(x\right) = 0\) to determine when the slope \(f\left(x\right)\) is \(0\):

\[\begin{align}f^\prime\left(x\right) &= 0\\ 3x^{2} - 30x + 75 &= 0 \\ x^{2} - 10x + 25 &= 0\ \\ \left(x - 5\right)^{2} &= 0 \\ x &= 5\end{align}\]

Substitute \(x = 5\) into \(f(x)\) to obtain the corresponding \(y\)-value

\(f\left(5\right) = -400\)

The point of inflection is \(\left(5, -400\right)\).

A graph of the cubic function f(x) = x^3 - 15x^2 + 75x - 525. The curve has a critical point at (5, -400), where the slope of the function is zero. On either side of this point the graph is increasing.

Substitute values on either side of \(x=5\) into \(f^{\prime}(x)\) to determine the nature of the stationary point

\[\begin{align}f^\prime\left(4\right) &= 3 \\ f^\prime\left(6\right) &= 3\end{align}\]

Substitute \(x = 5\) into \(f^{\prime\prime}(x)\) to confirm the stationary point is a stationary point of inflection

\[f^{\prime\prime}\left(5\right) = 0\]

A diagram illustrating a point is a point of inflection if the slope and therefore the first derivative is positive on the left of the stationary point and the slope and the first derivative is positive on the right of the stationary point.  

Example 2

Determine the stationary point of inflection in \(f\left(x\right) = x^{3} - 15x^{2} + 75x - 525\)

\[\begin{align} f\left(x\right) &= -x^{3} + 15x^{2} - 75x + 525 \\ f^\prime\left(x\right) &= -3x^{2} + 30x -75 \\ f^\prime\left(x\right) &= 0\\ -\left(3x^{2} - 30x + 75\right) &= 0 \\ -x^{2} + 10x -25 &= 0 \\ \left(x - 5\right)^{2} &= 0 \\ x &= 5 \\ f\left(5\right) &= 400 \end{align}  \]

The stationary point is at \(\left(5, 400\right)\).

To determine the nature of the stationary point we can substitute values either side of \(x=5\) into \(f^\prime(x)\)

\[ f^\prime\left(6\right) = 3 \] \[f^\prime\left(4\right) = 3\]

As the sign does not change, this is a stationary point of inflection.

A diagram illustrating a point is a point of inflection if the slope and therefore the first derivative is negative on the left of the stationary point and the slope and the first derivative is negative on the right of the stationary point.