In this note, you will learn:
· Maximum and minimum points with relations to first and second derivatives
Maximum and minimum points with relations to first and second derivatives
Looking the diagram shown above, we are able to clearly notice that there are certain points on the graph which are shown in blue, red and green.
All the points (blue, red and green) are called stationary points, which are points on the curve where the value of the derivative is zero. i.e. dy/dx = 0
However, when dy/dx = 0, it does not give a definite answer as to what it means. As you can see from the diagram shown above, there are three scenarios which would deem the value of dy/dx to be equal to zero:
· At points B and D where dy/dx = 0, the stationary points are called maximum points. As the curve passes through B and D, dy/dx changes from a positive value to a negative value.
· On the other hand, at points C and F where dy/dx = 0, the stationary points are called minimum points. As the curve passes through C and F, dy/dx changes from a negative value to a positive value.
However, there is one special case: At point G where dy/dx = 0, the curve passes through G and the value of dy/dx changes from a positive value to zero, before turning positive once again.
This type of stationary point is called point of inflexion. Points of inflexion can occur when:
i) dy/dx changes from a negative value to zero at the point of inflexion, before turning negative from zero once again.
ii) dy/dx changes from a positive value to zero at the point of inflexion, before turning positive from zero once again.
A visual representation of both i) and ii) are as shown below:
· First Derivative Test
Now that we know that stationary points occur when dy/dx = 0 and are presented in the form of turning points/ points of inflexion, we will be learning about how we can find the nature of stationary points next.
When a question asks for the “nature of a point”, what they basically want to know is if the point is a maximum, minimum or a point of inflexion. To find the nature of a point, we will have to use the First Derivative Test.
We will have to first find the derivative of the function, dy/dx, then solve for the values of x in order to find the coordinates of the stationary point(s) of the curve. After which, we will have to pick two random points that are close to the value of x: one less than the value of x, and one more than the value of x so that we can substitute it into the derivative to make a comparison to know if the stationary point is a maximum, minimum or a point of inflexion.
- If the gradient dy/dx changes from a positive value to a negative value as x increases through the stationary point, then the stationary point is a maximum point.
- If the gradient dy/dx changes from a negative value to a positive value as x increases through the stationary point, then stationary point is a minimum point.
- If the gradient dy/dx remains positive or negative as x increase through the stationary point, then the stationary point is a stationary point of inflexion.
Let’s take a look at how this works in the example down below:
Question 1: Find the coordinates of the stationary points on the curve y = (2/3) x3 – 8x + (2/3) and determine the nature of these points.
Solution:
y = (2/3) x^3 – 8x + (2/3)
dy/dx = 2x^2 – 8
For stationary points, dy/dx = 0.
2x^2 – 8 = 0
2x^2 = 8
x^2 = 4
x = 2 or -2
When x = 2, y = -10; and when x = -2, y = 34/3
Find the stationary point of the curve y = x^3 – 1 and determine its nature.
Solution:
y = x^3 – 1
dy/dx = 3x^2
For a stationary point, dy/dx = 0.
3x^2 = 0
x = 0
When x = 0, y = -1.
Therefore, the stationary point is (0, -1).
Therefore, (0, -1) is a point of inflexion.
· Second Derivative Test
We know that for the first derivative test, it will give us the positions of the stationary points and can only provide us with the nature of the stationary points on a curve with the help of information about the behavior of the first derivative before and after the stationary point. However, with the second derivative test, we can determine the nature of the stationary point without additional information!
As you can see from the diagram above:
· From A to the stationary point, gradient dy/dx is increasing from a negative value to zero.
· From the stationary point to B, gradient dy/dx is increasing from zero to a positive value.
Therefore, the second derivative test is basically using the value of the second derivative, or the rate of change of the first derivative, to determine the nature of the stationary point on the curve.
Hence,
· If dy/dx = 0 and d^2 y/dx^2 > 0 at a point, then the point is a minimum point.
(Rate of change of dy/dx is increasing constantly)
· If dy/dx = 0 and d^2 y/dx^2 < 0 at a point, then the point is a maximum point.
(Rate of change of dy/dx is decreasing constantly)
However, what if d^2 y/dx^2 = 0?
If d^2 y/dx^2 = 0, there is no additional information which we can derive from it, hence we must use the comparison method of values smaller and greater than the value of x to determine the nature of the stationary point.
Let’s take a look at how the second derivative test can be used in the example shown below:
The equation of a curve is y = x^3 + 2x^2 – 4x + 8.
i) Find the expression for dy/dx and obtain the coordinates of the stationary points of the curve.
ii) Find an expression for d^2 y/dx^2 and hence determine the nature of these stationary points.
Solution:
i) y = x^3 + 2x^2 – 4x + 8
dy/dx = 3x^2 + 4x – 4
For stationary points, dy/dx = 0.
3x^2 + 4x – 4 = 0
(3x – 2) (x + 2) = 0
x = 2/3 or -2
When x = 2/3, y = 176/27; and when x = -2, y = 16.
Therefore, the stationary points are (2/3, 176/27) and (-2, 16).
ii) d^2 y/dx^2 = 6x + 4
When x = 2/3, d^2 y/dx^2 = 8 > 0.
When x = -2, d^2 y/dx^2 = -8 < 0.
Therefore, (2/3, 6.52) is a minimum point and (-2, 16) is a maximum point.
And that’s all for today, students! Math Lobby hopes that after this article, you have a clear understanding on the topic of differentiation with the relations between the first and second derivatives and stationary points on a curve (maximum, minimum and points of inflexion)!
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