Lesson Explainer: Convexity and Points of Inflection Mathematics • Higher Education

In this explainer, we will learn how to determine the convexity of a function as well as its inflection points using its second derivative.

Before you start with this explainer, you should be confident finding the first and second derivatives of functions using the standard rules for differentiation. You should also be able to use the first derivative test to find the nature of critical points.

Now, before we begin looking at examples and a method of using the second derivative instead of the first derivative test, we will look at what it means for a graph to be convex up or convex down or to have a point of inflection. To do this, we will consider the graphs of three common functions.

Definition: Convexity and Inflection

From the diagrams, we can see that 𝑓(π‘₯)=π‘₯ is a good example of a function that is convex downward over its entire domain; it curves upward and the value of its slope is increasing over its entire domain. An alternative way to think about this is that if the graph of the function lies above all its tangents over some interval, the function is convex downward over that interval. Similarly, 𝑔(π‘₯)=βˆ’π‘₯ is an example of a function that is convex upward over its entire domain; the function curves downward and the value of the slope is always decreasing over this localized interval.

Considering the tangents to the curve once again, we see that if a graph of a function lies below all its tangents over some interval, then it is convex upward over that interval.

Looking at the functions 𝑓 and 𝑔, we also notice that the critical point on the graph of 𝑓(π‘₯)=π‘₯ is an absolute minimum; it is the lowest point of the curve over its entire domain. The critical point on the graph of 𝑔(π‘₯)=βˆ’π‘₯ is an absolute maximum; it is the highest point of the curve over its entire domain.

However, the function β„Ž(π‘₯)=π‘₯ demonstrates something a little bit different. The turning point at (0,0) is known as a point of inflection. This is characterized by the convexity changing from convex up to convex down (as in function β„Ž) or convex down to convex up.

Now that we have the definitions, let us look at how we would determine the nature of a critical point and therefore its convexity.

Consider the function 𝑓(π‘₯)=π‘₯; the gradient function, which is given by its first derivative, is 𝑓′(π‘₯)=2π‘₯.

We can use this function to evaluate the slope or gradient of the function at any given point. We can also use the first derivative test to check the nature of the critical point.

For example, the critical point of the function occurs at values of π‘₯ such that 𝑓′(π‘₯)=0. 2π‘₯=0,π‘₯=0.

The first derivative test tells us that the nature of the critical point can be established by finding the slope of the tangent to the curve to either side of this point.

The slope of the tangent to the curve of 𝑦=𝑓(π‘₯) at π‘₯=βˆ’1 can be found by substituting π‘₯=βˆ’1 into the gradient function: 𝑓′(βˆ’1)=βˆ’2.

Similarly, the slope of the tangent at π‘₯=1 is 𝑓′(1)=2.

Since the slope changes from negative to positive about the critical point, the graph must be convex downward.

But now, let us consider, in detail, what is happening to the derivative:

  • Prior to the critical point, it is negative.
  • At the critical point, it is zero.
  • After the critical point, it is positive.

For our function 𝑓, the value of 𝑓′(π‘₯) is increasing; in other words, the rate of change of 𝑓′(π‘₯) is greater than zero.

The rate of change of 𝑓′(π‘₯) is its derivative 𝑓′′(π‘₯), and 𝑓′′(π‘₯)>0.

We can, therefore, use this to determine convexity:

If 𝑓′′(π‘₯)>0 for all values of π‘₯ in an interval (𝐼), then the graph is convex down over this interval.

This is known as the second derivative test, since evaluating the second derivative at the critical point gives us information about the nature of the extrema and thus the convexity of the curve.

Next, consider 𝑔(π‘₯)=βˆ’π‘₯. We can see that just before the critical point the tangent has a positive slope and just after the critical point the slope is negative. This tells us that 𝑓′(π‘₯) is decreasing, which could also be expressed as 𝑓′′(π‘₯)<0.

So, if we evaluate the second derivative at the critical point and it is less than zero, then we can deduce that we have a local maximum. We can extend the idea to give us the following rule:

If 𝑓′′(π‘₯)<0 for all values of π‘₯ in an interval (𝐼), then the graph is convex up for all values in this interval.

We now have two rules to help us determine the convexity of a curve. What do we do in the case where 𝑓′′(π‘₯)=0?

If 𝑓′′(π‘₯)=0 or is not defined, then this could be a point of inflection. However, we must not assume that any point where 𝑓′′(π‘₯)=0 is a point of inflection. Instead, we must evaluate the second derivative on either side of our critical point and check that the convexity does indeed change from convex down to convex up, or vice versa.

Definition: Using the Second Derivative to Identify Convexity and Inflection

  • If 𝑓′′(π‘₯)>0 for all π‘₯ in 𝐼, then 𝑓 is convex downward on 𝐼.
  • If 𝑓′′(π‘₯)<0 for all π‘₯ in 𝐼, then 𝑓 is convex upward on 𝐼.
  • If 𝑓′′(π‘₯)=0 or is undefined, an inflection point may exist (although, alone, this does not guarantee an inflection point). To confirm that there is an inflection point, there must also be a change of convexity on either side of this point.

Note

A point of inflection can occur at a critical point, but this is not a necessity. Consider the graph of the function 𝑓(π‘₯)=3π‘₯+6π‘₯.

The convexity of the function changes from convex down to convex up at π‘₯=βˆ’23. This is a point of inflection but not a critical point.

We will now look at an example of how to calculate the intervals over which a polynomial function is convex down or convex up.

Example 1: Finding Intervals of Upward and Downward Convexity of a Polynomial

Determine the intervals on which the function 𝑓(π‘₯)=βˆ’4π‘₯+π‘₯ is convex up and down.

Answer

We know the following:

If 𝑓′′(π‘₯)>0 for all values of π‘₯ in an interval (𝐼), then the graph is convex down for all values of π‘₯ in this interval, and if 𝑓′′(π‘₯)<0 for all values of π‘₯ in an interval (𝐼), then the graph is convex up for all values of π‘₯ in this interval.

We will, therefore, need to find the second derivative of our function and use this to determine the intervals on which 𝑓′′(π‘₯)>0 and 𝑓′′(π‘₯)<0.

The first derivative, 𝑓′(π‘₯), is 𝑓′(π‘₯)=5Γ—ο€Ήβˆ’4π‘₯+3Γ—π‘₯=βˆ’20π‘₯+3π‘₯.οŠͺοŠͺ

Then, we find the second derivative by differentiating 𝑓′(π‘₯) with respect to π‘₯: 𝑓′′(π‘₯)=4Γ—ο€Ήβˆ’20π‘₯+2Γ—(3π‘₯)=βˆ’80π‘₯+6π‘₯.

Now that we have the second derivative, we can determine the intervals on which 𝑓′′(π‘₯)>0 and 𝑓′′(π‘₯)<0.

To achieve this, we will begin by setting the second derivative equal to zero and solving for π‘₯: βˆ’80π‘₯+6π‘₯=0.

To solve this equation, we can factor the left-hand side: 2π‘₯ο€Ήβˆ’40π‘₯+3=0.

From this, we can now solve for π‘₯ as we know that either 2π‘₯ must be equal to zero or the contents of the parentheses must be equal to zero: 2π‘₯=0π‘₯=0, or βˆ’40π‘₯+3=040π‘₯=3π‘₯=340π‘₯=Β±ο„ž340π‘₯=Β±ο„ž340.

At this point, we can rationalize the denominator by multiplying the numerator and denominator by √40: √3√40=√3√40√40√40=√3√4√1040=2√3040=√3020.

The solutions to 𝑓′′(π‘₯)=0 are π‘₯=0, π‘₯=√3020, or π‘₯=βˆ’βˆš3020. Next, we will sketch the curve of 𝑓′′(π‘₯) to help us decide where it is less than, greater than, or equal to zero.

This is a cubic graph with a negative coefficient of π‘₯, which has roots of βˆ’βˆš3020, √3020, and zero.

Marking the region where the output of the function is less than zero in orange and the region where the output is greater than zero in pink, we get

We can, therefore, say that 𝑓′′(π‘₯)>0; hence, the function is convex down, over the intervals ο€Ώβˆ’βˆž,βˆ’βˆš3020 and ο€Ώ0,√3020.

Similarly, 𝑓′′(π‘₯)<0, and therefore the function is convex upward, on the intervals ο€Ώβˆ’βˆš3020,0 and ο€Ώβˆš3020,βˆžο‹.

The function is convex down on ο€Ώβˆ’βˆž,βˆ’βˆš3020 and ο€Ώ0,√3020 and convex up on ο€Ώβˆ’βˆš3020,0 and ο€Ώβˆš3020,βˆžο‹.

In our first example, we established the convexity of the function by using the second derivative. In our next example, we will look at how to determine whether a function has any inflection points.

Example 2: Finding the Inflection Point on the Curve of a Quadratic Function If It Exists

Determine the inflection points of the curve 𝑦=π‘₯+2π‘₯βˆ’5.

Answer

We can recall the following:

If 𝑃 is an inflection point, then 𝑓′′(π‘₯)=0 (or is undefined) and the curve is continuous and changes from convex upward to downward, or vice versa, at 𝑃.

We can, therefore, begin by calculating an expression for the second derivative of the equation.

Note that since 𝑦 is a polynomial, we can infer that it is continuous and differentiable over its entire domain.

Firstly, we differentiate function 𝑦 with respect to π‘₯, which gives us dd𝑦π‘₯=2π‘₯+2.

The second derivative, ddοŠ¨οŠ¨π‘¦π‘₯=2, is a positive constant, which is independent of π‘₯, meaning that ddovertheentiredomainοŠ¨οŠ¨π‘¦π‘₯>0.

Hence, 𝑦 is convex down for all values of π‘₯.

In conclusion, since the convexity of the function never changes, we have shown that the curve 𝑦=π‘₯+2π‘₯βˆ’5 has no inflection points.

In the next question, we are going to demonstrate how to use the second derivative to find the point of inflection of a curve.

Example 3: Finding the Inflection Point of the Curve of a Polynomial Function

Find the inflection point on the graph of 𝑓(π‘₯)=π‘₯βˆ’9π‘₯+6π‘₯.

Answer

We can recall the following:

If 𝑃 is an inflection point, then 𝑓′′(π‘₯)=0 (or is undefined) and the curve is continuous and changes from convex upward to downward, or vice versa, at 𝑃.

As we are told that the graph has an inflection point, we will begin by finding the second derivative.

Firstly, we differentiate the function 𝑓(π‘₯) with respect to π‘₯, which gives us 𝑓′(π‘₯)=3π‘₯βˆ’2Γ—(9π‘₯)+6=3π‘₯βˆ’18π‘₯+6.

To find 𝑓′′(π‘₯), we differentiate 𝑓′(π‘₯): 𝑓′′(π‘₯)=6π‘₯βˆ’18.

It is worth noting that 𝑓′ is a quadratic function: it is continuous and differentiable over its entire domain, thereby guaranteeing that 𝑓′′ will exist for all real values of π‘₯.

We know that there could be a point of inflection when the second derivative is equal to zero, so we will set it equal to zero and solve for π‘₯: 6π‘₯βˆ’18=06π‘₯=18π‘₯=3.

However, 𝑓′′(π‘₯)=0 does not guarantee a point of inflection. We will, therefore, check the convexity of the curve on either side of the point π‘₯=3.To do this, we will check 𝑓′′(2) and 𝑓′′(4).

π‘₯𝑓′′(π‘₯)
26Γ—2βˆ’18=βˆ’6<0
30
46Γ—4βˆ’18=6>0

We can see that 𝑓′′(2)<0 and 𝑓′′(4)>0; therefore, the curve is going from convex up to convex down. We can conclude that a point of inflection occurs at π‘₯=3.

To find the 𝑦-coordinate of this inflection point, we will substitute π‘₯=3 into 𝑓(π‘₯): 𝑓(3)=3βˆ’9Γ—3+6Γ—3=βˆ’36.

The inflection point on the graph of 𝑓(π‘₯)=π‘₯βˆ’9π‘₯+6π‘₯ occurs at (3,βˆ’36).

In our next two examples, we will investigate how the standard rules of differentiation can also be applied to aid in testing for convexity and points of inflection, with a particular focus on trigonometric and logarithmic functions.

Example 4: Finding the Inflection Point of a Function Involving Trigonometric Functions in a Given Interval

Given that 𝑓(π‘₯)=4π‘₯+4π‘₯sincos, where 0≀π‘₯β‰€πœ‹2, determine the inflection points of 𝑓.

Answer

We can recall the following:

If 𝑃 is an inflection point, then 𝑓′′(π‘₯)=0 (or is undefined) and the curve is continuous and changes from convex upward to downward or, vice versa, at 𝑃.

Firstly, we differentiate the function 𝑓(π‘₯) with respect to π‘₯. To do this, we need to recall the following standard derivatives: ddsincosddcossinπ‘₯(π‘Žπ‘₯)=π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=βˆ’π‘Žπ‘Žπ‘₯.

Applying these to the individual terms in our function, 𝑓′(π‘₯)=44π‘₯βˆ’44π‘₯.cossin

Next, we differentiate 𝑓′(π‘₯) to find the second derivative: 𝑓′′(π‘₯)=βˆ’164π‘₯βˆ’164π‘₯.sincos

A point of inflection occurs when the second derivative is equal to zero (or does not exist) and when the convexity changes, so we set 𝑓′′(π‘₯)=0 and solve for π‘₯, remembering to restrict the set of solutions to the interval 0≀π‘₯β‰€πœ‹2.

Note

The function 𝑓′′(π‘₯) is the sum of two continuous function. This means it is, itself, continuous and so it is defined over its entire domain: 0=βˆ’164π‘₯βˆ’164π‘₯0=4π‘₯+4π‘₯βˆ’4π‘₯=4π‘₯βˆ’1=4π‘₯4π‘₯.sincossincoscossinsincos

At this stage, we can recall the trigonometric identity tansincosπ‘₯=π‘₯π‘₯.

Using tansincos4π‘₯=4π‘₯4π‘₯. we get βˆ’1=4π‘₯,tan which we can then solve for π‘₯: tan(βˆ’1)=4π‘₯βˆ’πœ‹4=4π‘₯.

At this point, we must remember that tanπ‘₯ is periodic with a period of πœ‹ radians, so this tells us that there might be more than one solution.

To find the possible solutions, we must consider our original interval; however, we are going to amend this by multiplying by 4 to give us 0≀4π‘₯≀2πœ‹.

We find all possible values for 4π‘₯ within our interval by adding multiples of πœ‹ to our solution to give us 4π‘₯=3πœ‹4,7πœ‹4.

Finally, we can divide through by 4 to find π‘₯: π‘₯=3πœ‹16,7πœ‹16.

We know that satisfying the criterion 𝑓′′(π‘₯)=0 does not guarantee a point of inflection. We must also check the convexity of the function on either side of the two values of π‘₯. Let us use π‘₯=0.5 and π‘₯=0.6 as values on either side of 3πœ‹16, and 1.3 and 1.4 on either side of 7πœ‹16.

π‘₯𝑓′′(π‘₯)
0.5βˆ’7.89…<0
0.60.99…>0
1.36.63…>0
1.4βˆ’2.30…<0

We can see that, about the point π‘₯=3πœ‹16, the graph changes from convex upward to convex downward and, about the point π‘₯=7πœ‹16, the graph changes from convex downward to convex upward. We can, therefore, conclude that a point of inflection occurs at π‘₯=3πœ‹16 and π‘₯=7πœ‹16.

To find the corresponding 𝑦-coordinates, we substitute each value of π‘₯ into the original function 𝑓(π‘₯): 𝑓3πœ‹16=0,𝑓7πœ‹16=0.

Given that 𝑓(π‘₯)=4π‘₯+4π‘₯sincos, where 0≀π‘₯β‰€πœ‹2, the inflection points of 𝑓 lie at ο€Ό3πœ‹16,0 and ο€Ό7πœ‹16,0.

In our final example, we will demonstrate how to apply this procedure for a function involving the natural logarithm.

Example 5: Finding the Point of Inflection, If It Exists, of a Function Involving a Logarithm

Find (if any) the inflection points of 𝑓(π‘₯)=3π‘₯2π‘₯ln.

Answer

If 𝑃 is an inflection point, then 𝑓′′(π‘₯)=0 (or is undefined) and the curve is continuous and changes from convex upward to downward, or vice versa, at 𝑃.

To find the points of inflection, we will evaluate the second derivative of our function and set it equal to zero.

On inspection of our function, we can see that it is the product of two functions: 3π‘₯2π‘₯.andln

Therefore, we will use the product rule to differentiate, which states that ddddddπ‘₯(𝑒𝑣)=𝑒𝑣π‘₯+𝑣𝑒π‘₯.

Let 𝑒=3π‘₯ and 𝑣=2π‘₯ln.

Then, we differentiate with respect to π‘₯ to give us ddanddd𝑒π‘₯=6π‘₯𝑣π‘₯=1π‘₯.

Using the product rule, we find 𝑓′(π‘₯)=3π‘₯Γ—1π‘₯+6π‘₯Γ—2π‘₯=3π‘₯+6π‘₯2π‘₯.lnln

To find the second derivative, we will once again use the product rule to find the derivative of 6π‘₯2π‘₯ln: 𝑓′′(π‘₯)=3+6+62π‘₯=9+62π‘₯.lnln

Now that we have the second derivative, we can set it equal to zero and solve for π‘₯: 0=9+62π‘₯βˆ’9=62π‘₯βˆ’32=2π‘₯𝑒=2π‘₯12𝑒=π‘₯.lnlnlnοŽͺοŽͺ

We can see that there is potentially an inflection point at π‘₯=12𝑒.οŽͺ

However, 𝑓′′(π‘₯)=0 does not guarantee the existence of a point of inflection, and so we will check the values of 𝑓′′(π‘₯) on either side of this. If we calculate 12𝑒οŽͺ, it is approximately 0.112; therefore, we can consider π‘₯ values of 0.1 and 0.12.

π‘₯𝑓′′(π‘₯)
0.1βˆ’0.656…<0
0.120.437…>0

We can see from the table that 𝑓′′(0.1)<0 and 𝑓′′(0.12)>0. This tells us that the curve goes from being convex up to being convex down. This confirms the existence of a point of inflection at π‘₯=12𝑒.οŽͺ

We can now substitute this value of π‘₯ into our original function to find the 𝑦-coordinate of our point of inflection: 𝑓12𝑒=3𝑒2×2𝑒2=3𝑒4Γ—βˆ’32𝑒=34π‘’Γ—βˆ’32=βˆ’98𝑒.οŽͺοŽͺοŽͺlnln

So, we can conclude that the inflection point of 𝑓(π‘₯)=3π‘₯2π‘₯ln is at 𝑒2,98𝑒οŽͺ.

We can apply what we have learned about convexity to parametric equations. First, we will need to recall how we can find the first and second derivatives of parametric equations.

Definition: Derivative of a Parametric Equation

Let 𝑓 and 𝑔 be differentiable functions, such that π‘₯ and 𝑦 are a pair of parametric equations: π‘₯=𝑓(𝑑),𝑦=𝑔(𝑑).

Then, we can define the derivative of 𝑦 with respect to π‘₯ as dd𝑦π‘₯=,ddddο˜οο—ο when ddπ‘₯𝑑≠0.

Definition: Second Derivative of a Parametric Equation

Let 𝑓 and 𝑔 be differentiable functions, such that π‘₯ and 𝑦 are a pair of parametric equations: π‘₯=𝑓(𝑑),𝑦=𝑔(𝑑).

Then, we can define the second derivative of 𝑦 with respect to π‘₯ as ddοŠ¨οŠ¨οο˜ο—ο—οπ‘¦π‘₯=,dddddd when ddπ‘₯𝑑≠0.

Now let us look at an example of how we can use this.

Example 6: Determining the Convexity of a Parametric Curve at a Given Point

Consider the parametric curve π‘₯=πœƒcos and 𝑦=πœƒsin. Determine whether this curve is convex up, convex down, or neither at πœƒ=πœ‹6.

Answer

We have been asked about the convexity of a curve, so we will need to evaluate the second derivative of the curve at the given point. Since this is a parametric curve, we can use the following formula to find its second derivative: ddοŠ¨οŠ¨οΌο˜ο—ο—οΌπ‘¦π‘₯=.dddddd

Let us start by finding ddπ‘₯πœƒ and ddπ‘¦πœƒ. We can do this by using trigonometric differentiation. We have that ddsinπ‘₯πœƒ=βˆ’πœƒ and ddcosπ‘¦πœƒ=πœƒ.

So, we find that ddcossincot𝑦π‘₯=πœƒβˆ’πœƒ=βˆ’πœƒ.

Now we can differentiate dd𝑦π‘₯ with respect to πœƒ. Using the differentials of reciprocal trigonometric functions, we have that ddddcscπœƒο€½π‘¦π‘₯=πœƒ.

Now we are able to substitute these into the formula for the second derivative. In doing this, we obtain ddcscsincscοŠ¨οŠ¨οŠ¨οŠ©π‘¦π‘₯=πœƒβˆ’πœƒ=βˆ’πœƒ.

Now that we have found the second derivative of the curve, we need to substitute in the value of the parameter at the point at which we are trying to find the convexity. The given point is πœƒ=πœ‹6, so we substitute this into our equation, which gives us ddcscοŠ¨οŠ¨οΌοŠ²οŠ©π‘¦π‘₯|||=βˆ’πœ‹6=βˆ’8.ο‘½οŽ₯

We can see that the second derivative at πœƒ=πœ‹6 is negative, so we have that ddοŠ¨οŠ¨οΌοŠ²π‘¦π‘₯|||<0.ο‘½οŽ₯

Therefore, our function must be convex up at this point.

We will finish by recapping some key points of this explainer.

Key Points

  • If 𝑓′′(π‘₯)>0 for all π‘₯ in 𝐼, then 𝑓 is convex downward on 𝐼.
  • If 𝑓′′(π‘₯)<0 for all π‘₯ in 𝐼, then 𝑓 is convex upward on 𝐼.
  • An inflection point occurs when the convexity of the graph changes and 𝑓′′(π‘₯)=0 or is undefined (although, alone, this does not guarantee an inflection point).

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