Lesson Explainer: Concavity and Points of Inflection

In this explainer, we will learn how to determine the concavity 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 concave up or concave down or to have a point of inflection. To do this, we will consider the graphs of three common functions.

Definition: Concavity and Inflection

From the diagrams, we can see that 𝑓(𝑥)=𝑥 is a good example of a function that is concave upward 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 concave upward over that interval. Similarly, 𝑔(𝑥)=𝑥 is an example of a function that is concave downward 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 concave downward over that interval.

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

However, the function (𝑥)=𝑥 demonstrates something a little bit different. The turning point at (0,0) is known as a point of inflection. This is characterized by the concavity changing from concave down to concave up (as in function ) or concave up to concave down.

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

Consider the function 𝑓(𝑥)=𝑥; the gradient function, which is given by its first derivative, is 𝑓(𝑥)=2𝑥.

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 𝑥 such that 𝑓(𝑥)=0. 2𝑥=0,𝑥=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(𝑥) at 𝑥=1 can be found by substituting 𝑥=1 into the gradient function: 𝑓(1)=2.

Similarly, the slope of the tangent at 𝑥=1 is 𝑓(1)=2.

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

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(𝑥) is increasing; in other words, the rate of change of 𝑓(𝑥) is greater than zero.

The rate of change of 𝑓(𝑥) is its derivative 𝑓(𝑥), and 𝑓(𝑥)>0.

We can, therefore, use this to determine concavity:

If 𝑓(𝑥)>0 for all values of 𝑥 in an interval (𝐼), then the graph is concave up 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 concavity of the curve.

Next, consider 𝑔(𝑥)=𝑥. 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(𝑥) is decreasing, which could also be expressed as 𝑓(𝑥)<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(𝑥)<0 for all values of 𝑥 in an interval (𝐼), then the graph is concave down for all values in this interval.

We now have two rules to help us determine the concavity of a curve. What do we do in the case where 𝑓(𝑥)=0?

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

Definition: Using the Second Derivative to Identify Concavity and Inflection

  • If 𝑓(𝑥)>0 for all 𝑥 in 𝐼, then 𝑓 is concave upward on 𝐼.
  • If 𝑓(𝑥)<0 for all 𝑥 in 𝐼, then 𝑓 is concave downward on 𝐼.
  • If 𝑓(𝑥)=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 concavity 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(𝑥)=3𝑥+6𝑥.

The concavity of the function changes from concave up to concave down at 𝑥=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 concave up or concave down.

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

Determine the intervals on which the function 𝑓(𝑥)=4𝑥+𝑥 is concave up and down.

Answer

We know the following:

If 𝑓(𝑥)>0 for all values of 𝑥 in an interval (𝐼), then the graph is concave up for all values of 𝑥 in this interval, and if 𝑓(𝑥)<0 for all values of 𝑥 in an interval (𝐼), then the graph is concave down for all values of 𝑥 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(𝑥)>0 and 𝑓(𝑥)<0.

The first derivative, 𝑓(𝑥), is 𝑓(𝑥)=5×4𝑥+3×𝑥=20𝑥+3𝑥.

Then, we find the second derivative by differentiating 𝑓(𝑥) with respect to 𝑥: 𝑓(𝑥)=4×20𝑥+2×(3𝑥)=80𝑥+6𝑥.

Now that we have the second derivative, we can determine the intervals on which 𝑓(𝑥)>0 and 𝑓(𝑥)<0.

To achieve this, we will begin by setting the second derivative equal to zero and solving for 𝑥: 80𝑥+6𝑥=0.

To solve this equation, we can factor the left-hand side: 2𝑥40𝑥+3=0.

From this, we can now solve for 𝑥 as we know that either 2𝑥 must be equal to zero or the contents of the parentheses must be equal to zero: 2𝑥=0𝑥=0, or 40𝑥+3=040𝑥=3𝑥=340𝑥=±340𝑥=±340.

At this point, we can rationalize the denominator by multiplying the numerator and denominator by 40: 340=3404040=341040=23040=3020.

The solutions to 𝑓(𝑥)=0 are 𝑥=0, 𝑥=3020, or 𝑥=3020. Next, we will sketch the curve of 𝑓(𝑥) 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 𝑥, which has roots of 3020, 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(𝑥)>0; hence, the function is concave up, over the intervals ,3020 and 0,3020.

Similarly, 𝑓(𝑥)<0, and therefore the function is concave downward, on the intervals 3020,0 and 3020,.

The function is concave up on ,3020 and 0,3020 and concave down on 3020,0 and 3020,.

In our first example, we established the concavity 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=𝑥+2𝑥5.

Answer

We can recall the following:

If 𝑃 is an inflection point, then 𝑓(𝑥)=0 (or is undefined) and the curve is continuous and changes from concave 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 𝑥, which gives us dd𝑦𝑥=2𝑥+2.

The second derivative, dd𝑦𝑥=2, is a positive constant, which is independent of 𝑥, meaning that ddovertheentiredomain𝑦𝑥>0.

Hence, 𝑦 is concave upward for all values of 𝑥.

In conclusion, since the concavity of the function never changes, we have shown that the curve 𝑦=𝑥+2𝑥5 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(𝑥)=𝑥9𝑥+6𝑥.

Answer

We can recall the following:

If 𝑃 is an inflection point, then 𝑓(𝑥)=0 (or is undefined) and the curve is continuous and changes from concave 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(𝑥) with respect to 𝑥, which gives us 𝑓(𝑥)=3𝑥2×(9𝑥)+6=3𝑥18𝑥+6.

To find 𝑓(𝑥), we differentiate 𝑓(𝑥): 𝑓(𝑥)=6𝑥18.

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 𝑥.

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 𝑥: 6𝑥18=06𝑥=18𝑥=3.

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

𝑥𝑓(𝑥)
26×218=6<0
30
46×418=6>0

We can see that 𝑓(2)<0 and 𝑓(4)>0; therefore, the curve is going from concave downward to concave upward. We can conclude that a point of inflection occurs at 𝑥=3.

To find the 𝑦-coordinate of this inflection point, we will substitute 𝑥=3 into 𝑓(𝑥): 𝑓(3)=39×3+6×3=36.

The inflection point on the graph of 𝑓(𝑥)=𝑥9𝑥+6𝑥 occurs at (3,36).

In our next two examples, we will investigate how the standard rules of differentiation can also be applied to aid in testing for concavity 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(𝑥)=4𝑥+4𝑥sincos, where 0𝑥𝜋2, determine the inflection points of 𝑓.

Answer

We can recall the following:

If 𝑃 is an inflection point, then 𝑓(𝑥)=0 (or is undefined) and the curve is continuous and changes from concave upward to downward or, vice versa, at 𝑃.

Firstly, we differentiate the function 𝑓(𝑥) with respect to 𝑥. To do this, we need to recall the following standard derivatives: ddsincosddcossin𝑥(𝑎𝑥)=𝑎𝑎𝑥,𝑥(𝑎𝑥)=𝑎𝑎𝑥.

Applying these to the individual terms in our function, 𝑓(𝑥)=44𝑥44𝑥.cossin

Next, we differentiate 𝑓(𝑥) to find the second derivative: 𝑓(𝑥)=164𝑥164𝑥.sincos

A point of inflection occurs when the second derivative is equal to zero (or does not exist) and when the concavity changes, so we set 𝑓(𝑥)=0 and solve for 𝑥, remembering to restrict the set of solutions to the interval 0𝑥𝜋2.

Note

The function 𝑓(𝑥) is the sum of two continuous function. This means it is, itself, continuous and so it is defined over its entire domain: 0=164𝑥164𝑥0=4𝑥+4𝑥4𝑥=4𝑥1=4𝑥4𝑥.sincossincoscossinsincos

At this stage, we can recall the trigonometric identity tansincos𝑥=𝑥𝑥.

Using tansincos4𝑥=4𝑥4𝑥. we get 1=4𝑥,tan which we can then solve for 𝑥: tan(1)=4𝑥𝜋4=4𝑥.

At this point, we must remember that tan𝑥 is periodic with a period of 𝜋 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 04𝑥2𝜋.

We find all possible values for 4𝑥 within our interval by adding multiples of 𝜋 to our solution to give us 4𝑥=3𝜋4,7𝜋4.

Finally, we can divide through by 4 to find 𝑥: 𝑥=3𝜋16,7𝜋16.

We know that satisfying the criterion 𝑓(𝑥)=0 does not guarantee a point of inflection. We must also check the concavity of the function on either side of the two values of 𝑥. Let us use 𝑥=0.5 and 𝑥=0.6 as values on either side of 3𝜋16, and 1.3 and 1.4 on either side of 7𝜋16.

𝑥𝑓(𝑥)
0.57.89<0
0.60.99>0
1.36.63>0
1.42.30<0

We can see that, about the point 𝑥=3𝜋16, the graph changes from concave downward to concave upward and, about the point 𝑥=7𝜋16, the graph changes from concave upward to concave downward. We can, therefore, conclude that a point of inflection occurs at 𝑥=3𝜋16 and 𝑥=7𝜋16.

To find the corresponding 𝑦-coordinates, we substitute each value of 𝑥 into the original function 𝑓(𝑥): 𝑓3𝜋16=0,𝑓7𝜋16=0.

Given that 𝑓(𝑥)=4𝑥+4𝑥sincos, where 0𝑥𝜋2, the inflection points of 𝑓 lie at 3𝜋16,0 and 7𝜋16,0.

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(𝑥)=3𝑥2𝑥ln.

Answer

If 𝑃 is an inflection point, then 𝑓(𝑥)=0 (or is undefined) and the curve is continuous and changes from concave 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: 3𝑥2𝑥.andln

Therefore, we will use the product rule to differentiate, which states that dddddd𝑥(𝑢𝑣)=𝑢𝑣𝑥+𝑣𝑢𝑥.

Let 𝑢=3𝑥 and 𝑣=2𝑥ln.

Then, we differentiate with respect to 𝑥 to give us ddanddd𝑢𝑥=6𝑥𝑣𝑥=1𝑥.

Using the product rule, we find 𝑓(𝑥)=3𝑥×1𝑥+6𝑥×2𝑥=3𝑥+6𝑥2𝑥.lnln

To find the second derivative, we will once again use the product rule to find the derivative of 6𝑥2𝑥ln: 𝑓(𝑥)=3+6+62𝑥=9+62𝑥.lnln

Now that we have the second derivative, we can set it equal to zero and solve for 𝑥: 0=9+62𝑥9=62𝑥32=2𝑥𝑒=2𝑥12𝑒=𝑥.lnlnln

We can see that there is potentially an inflection point at 𝑥=12𝑒.

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

𝑥𝑓(𝑥)
0.10.656<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 concave down to being concave up. This confirms the existence of a point of inflection at 𝑥=12𝑒.

We can now substitute this value of 𝑥 into our original function to find the 𝑦-coordinate of our point of inflection: 𝑓12𝑒=3𝑒2×2𝑒2=3𝑒4×32𝑒=34𝑒×32=98𝑒.lnln

So, we can conclude that the inflection point of 𝑓(𝑥)=3𝑥2𝑥ln is at 𝑒2,98𝑒.

We will finish by recapping the key points from this explainer.

Key Points

  • If 𝑓(𝑥)>0 for all 𝑥 in 𝐼, then 𝑓 is concave upward on 𝐼.
  • If 𝑓(𝑥)<0 for all 𝑥 in 𝐼, then 𝑓 is concave downward on 𝐼.
  • An inflection point occurs when the concavity of the graph changes and 𝑓(𝑥)=0 or is undefined (although, alone, this does not guarantee an inflection point).

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.