## Stationary Points

Contents

## Optional Investigation

Complete the table below for the quadratic function \(f(x)\):

\begin{align*} f(x) &= x^{2} + 2x + 1 \\ f'(x) &= \ldots \ldots \ldots \end{align*}

\(x\)-value \(-\text{3}\) \(-\text{1}\) \(\text{0}\) \(\text{1}\) \(\text{3}\) Gradient of \(f\) Sign of gradient \(\begin{array}{c@{\;}c@{\;}l} \text{Increasing function } (\nearrow) & & \\ \text{Decreasing function } (\searrow) & & \\ \text{Maximum TP } (\cap) && \\ \text{Minimum TP } (\cup) && \end{array}\) - Use the table to draw a rough sketch of the graph of \(f(x)\).
- Solve for \(x\) if \(f'(x) = 0\).
- Indicate solutions to \(f'(x) = 0\) on the graph.
Complete the table below for the cubic function \(g(x)\):

\begin{align*} g(x) &= 2x^{3} + 3x^{2} -12x \\ g'(x) &= \ldots \ldots \ldots \end{align*}

\(x\)-value \(-\text{3}\) \(-\text{2}\) \(\text{0}\) \(\text{1}\) \(\text{3}\) Gradient of \(g\) Sign of gradient \(\begin{array}{c@{\;}c@{\;}l} \text{Increasing function } (\nearrow) & & \\ \text{Decreasing function } (\searrow) & & \\ \text{Maximum TP } (\cap) && \\ \text{Minimum TP } (\cup) && \end{array}\) - Use the table to draw a rough sketch of the graph of \(g(x)\).
- Solve for \(x\) if \(g'(x) = 0\).
- Indicate solutions to \(g'(x) = 0\) on the graph.
Complete the following sentence:

The derivative describes the \(\ldots\ldots\) of a tangent to a curve at a given point and we have seen that the \(\ldots\ldots\) of a curve at its stationary point(s) is equal to \(\ldots\ldots\). Therefore, we can use \(\ldots\ldots\) as a tool for finding the stationary points of the graphs of quadratic and cubic functions.

To determine the coordinates of the stationary point(s) of \(f(x)\):

- Determine the derivative \(f'(x)\).
- Let \(f'(x) = 0\) and solve for the \(x\)-coordinate(s) of the stationary point(s).
- Substitute value(s) of \(x\) into \(f(x)\) to calculate the \(y\)-coordinate(s) of the stationary point(s).

## Example

### Question

Calculate the stationary points of the graph of \(p(x)= {x}^{3} – 6{x}^{2} + 9x – 4\).

### Determine the derivative of \(p(x)\)

Using the rules of differentiation we get:

\[{p}'(x)=3{x}^{2} – 12x + 9\]

### Let \({p}'(x)=0\) and solve for \(x\)

\begin{align*} 3{x}^{2} – 12x + 9 & = 0 \\ {x}^{2}-4x+3 & = 0 \\ (x-3)(x-1) & = 0 \\ \therefore x = 1 & \text{ or } x = 3 \end{align*}

Therefore, the \(x\)-coordinates of the turning points are \(x=1\) and \(x=3\).

### Substitute the \(x\)-values into \(p(x)\)

We use the \(x\)-coordinates to calculate the corresponding \(y\)-coordinates of the stationary points.

\begin{align*} p(1) & = {(1)}^{3}-6{(1)}^{2} + 9(1)-4 \\ & = 1 – 6 + 9 – 4\\ & = 0 \end{align*}\begin{align*} p(3) & = {(3)}^{3}- 6{(3)}^{2} + 9(3)-4 \\ & = 27 – 54 + 27 – 4 \\ & = -4 \end{align*}

### Write final answer

The turning points of the graph of \(p(x)= {x}^{3} – 6{x}^{2} + 9x – 4\) are \((1;0)\) and \((3;-4)\).

**Local maximum and local minimum**

We have seen that the graph of a quadratic function can have either a minimum turning point (“smile”) or a maximum turning point (“frown”).

For cubic functions, we refer to the turning (or stationary) points of the graph as local minimum or local maximum turning points. The diagram below shows local minimum turning point \(A(1;0)\) and local maximum turning point \(B(3;4)\). These points are described as a local (or relative) minimum and a local maximum because there are other points on the graph with lower and higher function values.