3.03 The second derivative and properties of graphs

The first derivative and properties of graphs

Let's review what the first derivative can tell us about a graph and then look further at second derivatives.

 

Increasing and decreasing

When a function is increasing, that is as the $x$x values increase the $y$y values increase, this relates to a positive gradient and hence, $f'\left(x\right)>0$f′(x)>0. 

When a function is decreasing, that is as the $x$x values increase the $y$y values decrease, this relates to a negative gradient and hence, $f'\left(x\right)<0$f′(x)<0. 

For the graphs below observe where the function is increasing (indicated in green sections), where it is decreasing decreasing (indicated in blue) and where the derivative function is located.

 

Stationary points

A stationary point is where the derivative is zero, that is a point $\left(a,f(a)\right)$(a,f(a)) is said to be a stationary point if $f'\left(a\right)=0$f′(a)=0. At this point the tangent is horizontal and thus, the instantaneous rate of change is zero - so the function is momentarily stationary. We have three types of stationary points:

• Local minimum: at this point $f\left(x\right)$f(x) has the lowest value in the region of this point. The graph changes from decreasing to increasing at this point. Hence, the first derivative will change from negative to positive.
• Local maximum: at this point $f\left(x\right)$f(x) has the highest value in the region of this point. The graph changes from increasing to decreasing at this point. Hence, the first derivative will change from positive to negative.

• Stationary point of inflection: at this point the gradient does not change either side of the point, it can be either positive either side of the stationary point or negative either side. At points of inflection the rate of change switches from increasing to decreasing or vice versa. So unlike the maximum or minimum it is not the graph itself changing between increasing and decreasing but the gradient of the graph. So the gradient changes from becoming less steep to becoming steeper or vice versa. 

Local maximum and minimum are also referred to as turning points, since the function changes between increasing and decreasing at these points.

The reason for the emphasis on "local" minimum or maximum is that they may give the minimum or maximum value within a region of the point but not over the entire domain of the graph. If a point gives the maximum or minimum value over the entire domain of a graph we refer to the point as a global maximum or global minimum. In the first graph below we have both a global and local minimum, as well as a local maximum, however, as the graph is unbounded there is no global maximum. The second graph below shows a restricted domain where the global maximum is in fact at an end-point and not a turning point.

For many functions a local maximum or minimum is also a global maximum or minimum, such as turning points for an unrestricted quadratic function.

 

The second derivative and properties of graphs

Just as the first derivative gives the rate of change of a function, the second derivative gives us the rate of change of the gradient function.

On an interval if:

The gradient is increasing	$\rightarrow$→	$f'\left(x\right)$f′(x) is increasing	$\rightarrow$→	$f''\left(x\right)>0$f′′(x)>0
The gradient is decreasing	$\rightarrow$→	$f'\left(x\right)$f′(x) is decreasing	$\rightarrow$→	$f''\left(x\right)<0$f′′(x)<0

Increasing gradient	Decreasing gradient

 

Concavity

We classify regions of a function as concave up where the gradient is increasing and concave down where the gradient is decreasing. As seen in the diagrams above this relates to the shape of the graph and can be tested using the second derivative.

What can happen when $f''\left(x\right)=0$f′′(x)=0? Recall, when the first derivative equals zero the graph is at a stationary point, the graph will show a key feature of either a turning point - a maximum or minimum where the graph changes between increasing and decreasing, or we see a stationary point of inflection where the graph is momentarily stationary but does not change between increasing and decreasing. 

Similarly, when the second derivative equals zero the gradient function is at a stationary point and the possibilities are:

• The gradient function is at a maximum and changes from increasing to decreasing. This means the graph of the original function is changing from concave up to concave down and the point will be a point of inflection.
• The gradient function is at a minimum and changes from decreasing to increasing. This means the graph of the original function is changing from concave down to concave up and the point will be a point of inflection.

• The gradient function is momentarily stationary but the graph does not change concavity through the point. This means the original graph will have a maximum if the gradient remains decreasing either side of the point or minimum if the gradient remains increasing either side of the point .

 

Worked examples

Example 1

Find where the function $f\left(x\right)=2x^3-9x^2-108x+2$f(x)=2x3−9x2−108x+2 is concave up and concave down.

Think: Concavity is determined according to the sign of the second derivative. 

Do: Find the second derivative:

$f'\left(x\right)$f′(x)	$=$=	$6x^2-18x-108$6x2−18x−108
$f''\left(x\right)$f′′(x)	$=$=	$12x-18$12x−18

To find where the curve is concave down solve  $f''\left(x\right)<0$f′′(x)<0:

$12x-18$12x−18	$<$<	$0$0
$12x$12x	$<$<	$18$18
$x$x	$<$<	$\frac{3}{2}$32​

To find where the curve is concave up we solve  $f''\left(x\right)>0$f′′(x)>0:

$12x-18$12x−18	$>$>	$0$0
$12x$12x	$>$>	$18$18
$x$x	$>$>	$\frac{3}{2}$32​

So the curve is concave down for $x<\frac{3}{2}$x<32​ and concave up for $x>\frac{3}{2}$x>32​.

 

Points of inflection

A point of inflection identifies a point on the curve where the graph changes concavity.

There is a point of inflection at $x=a$x=a if $f''\left(a\right)=0$f′′(a)=0 and the concavity changes at this point. There are two types of points of inflection:

• Stationary points of inflection: where the tangent to to curve at the point is horizontal and hence,$f'\left(a\right)=0$f′(a)=0.
• Non-stationary (ordinary) points of inflection: where tangent to the curve is not horizontal, $f'\left(a\right)\ne0$f′(a)≠0.

Here are two graphs together with their concavity and points of inflection highlighted. 

Stationary point of inflection	Two ordinary points of inflection

 

Classifying minimum, maximum and inflection points

Previously we have seen we can use the first derivative to find stationary points by solving $f'\left(x\right)=0$f′(x)=0 and classify them by testing the gradient just either side of a critical point. The second derivative now offers another way to test the nature of stationary points by checking the shape of the graph (concavity) at the point.

Since a minimum has a concave up shape, we can confirm a stationary point is a minimum by checking that $f''\left(x\right)>0$f′′(x)>0 at the point. Similarly, we can confirm a stationary point is a maximum by confirming the telltale concave down shape through showing $f''\left(x\right)<0$f′′(x)<0. If at a stationary point $f''\left(x\right)=0$f′′(x)=0 as well, we have identified a possible stationary point of inflection. To confirm the point is a point of inflection we must check that the concavity also changes at this point.

To find all possible points of inflection we can solve for when $f''\left(x\right)=0$f′′(x)=0. However, to confirm the point is a point of inflection we must check that the concavity also changes at this point. This can be done by checking the value of the second derivative just either side of the point. If the sign of the second derivative changes, we do indeed have a point of inflection.

If $f'\left(x\right)=0$f′(x)=0 and $f''\left(x\right)=0$f′′(x)=0 but the function does not change concavity through the point. This means the original graph will have a minimum if the graph is concave up, $f''\left(x\right)>0$f′′(x)>0, either side of the point or a maximum if the graph is concave down, $f''\left(x\right)<0$f′′(x)<0, either side of the point .

 

Worked examples

Example 2

Find and classify all stationary points on the curve $f\left(x\right)=x^4-2x^2+3$f(x)=x4−2x2+3 .

The first derivative is:

$f'\left(x\right)$f′(x)	$=$=	$4x^3-4x$4x3−4x
$f'\left(x\right)$f′(x)	$=$=	$4x\left(x^2-1\right)$4x(x2−1)
$f'\left(x\right)$f′(x)	$=$=	$4x\left(x+1\right)\left(x-1\right)$4x(x+1)(x−1)

 

 

Make $f'\left(x\right)=0$f′(x)=0 to find stationary points:  

$f'\left(x\right)$f′(x)	$=$=	$0$0
$4x\left(x+1\right)\left(x-1\right)$4x(x+1)(x−1)	$=$=	$0$0
$x$x	$=$=	$-1,0,1$−1,0,1

 

We can see that there are stationary points at $x=-1,0,1$x=−1,0,1. We now need to calculate the second derivative for each of these values to determine the nature of the stationary points.

$x$x	$-1$−1	$0$0	$1$1
$f''\left(x\right)$f′′(x)	$8$8	$-4$−4	$8$8
$Concavity$Concavity	$\smile$⌣	$\frown$⌢	$\smile$⌣
Meaning	positive concave up	negative concave down	positive concave up
Turning point classification	minimum	maximum	minimum

 

Graph verification.  Let's check if our findings are confirmed with a graph of the function.

All that's left now is to fully state the points and their classification:

At $x=-1$x=−1  $f''\left(x\right)=8$f′′(x)=8, hence the curve is concave up at this point. Therefore $\left(-1,2\right)$(−1,2) is a minimum turning point.

At $x=0$x=0  $f''\left(x\right)=-4$f′′(x)=−4, hence the curve is concave down at this point. Therefore $\left(0,3\right)$(0,3) is a maximum turning point.

At $x=1$x=1  $f''\left(x\right)=8$f′′(x)=8, hence the curve is concave up at this point. Therefore $\left(1,2\right)$(1,2) is a minimum turning point.

Example 3

Find any possible points of inflection on the curve $y=x^6$y=x6

Taking the first and second derivatives:

$y'$y′	$=$=	$6x^5$6x5
$y''$y′′	$=$=	$30x^4$30x4

 

As possible points of inflection occur where $f''\left(x\right)=0$f′′(x)=0 we can see there is a possible point of inflection at $x=0$x=0.

We will now check the second derivative either side of this point to check concavity and determine its nature:

$x$x	$-1$−1	$0$0	$1$1
$f''\left(x\right)$f′′(x)	$30$30	$0$0	$30$30
$Concavity$Concavity	$\smile$⌣	$.$.	$\smile$⌣
Meaning	positive concave up	stationary minimum turning point	positive concave up

As shown in the table, as there is no change in concavity at $x=0$x=0 and the graph is concave up either side of the point. So here we have a minimum turning point and not a point of inflection.

This example highlights the importance of checking concavity either side of points where $f''\left(x\right)=0$f′′(x)=0 . The first derivative of this function told us that there was a stationary point at $x=0$x=0. However, in this case, we still needed to look at the sign of the second derivative left and right of the point to determine if there was a change in concavity. 

 

Summary

The following table summarises our first derivative and second derivative curve analysis.

	$f'(x)<0$f′(x)<0 Decreasing graph	$f'(x)=0$f′(x)=0 Flat Spot	$f'(x)>0$f′(x)>0 Increasing graph
$f''(x)<0$f′′(x)<0 Concave down
$f''(x)=0$f′′(x)=0 Possible POI		or
$f''(x)>0$f′′(x)>0 Concave up

 

Practice questions

Question 1

Question 2

Question 3

Question 4