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:
\(y = x^2 - 2x - 8\)
\(y = -x^2-6x-8\)
\(y = 2x^3 - 12x^2 - 30x - 10\)
\(y = -2x^3 + 3x^2 + 36x - 6\)
\(y = x^3+3x^2+3x-2\)
Note: this exercise can be downloaded as a worksheet to practice with: worksheet
Answers Without Working
We find the derivative to be \(\frac{dy}{dx} = 2x-2\) and this curve has one stationary point:
\[\begin{pmatrix} 1,-9\end{pmatrix}\]
We find the derivative to be \(\frac{dy}{dx} = -2x-6\) and this curve has one stationary point:
\[\begin{pmatrix} -3,1\end{pmatrix}\]
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
\[\begin{pmatrix} 5,-210\end{pmatrix}\]
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
\[\begin{pmatrix} 3,75\end{pmatrix}\]
We find the derivative to be \(\frac{dy}{dx} = x^3+3x^2+3x-2\) and this curve has one stationary point:
\[\begin{pmatrix} -1,-3\end{pmatrix}\]
Answers with Working
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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:
\( y = 2x + \frac{8}{x}\)
\( y = -x - \frac{1}{x}\)
\( y = 3x + \frac{27}{x}\)
\( y = -2x - \frac{72}{x}\)
\( y = x + \frac{25}{x}\)
Note: this exercise can be downloaded as a worksheet to practice with: worksheet
Answers Without Working
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
\[\begin{pmatrix} 2,8\end{pmatrix}\]
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
\[\begin{pmatrix} 1,-2\end{pmatrix}\]
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
\[\begin{pmatrix} 3,18\end{pmatrix}\]
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
\[\begin{pmatrix} 6,-48\end{pmatrix}\]
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
\[\begin{pmatrix} 5,10\end{pmatrix}\]
Answers with Working
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