Lesson Explainer: Interpreting Graphs of Derivatives Mathematics • Higher Education

In this explainer, we will learn how to connect a function to the graphs of its first and second derivatives.

The derivatives of a function give us many different techniques for describing the different properties of a curve. For example, the slope of a curve is represented by its first derivative and the concavity of the curve is represented by its second derivative.

For a differentiable function 𝑓, we know that the local extrema will occur when its derivative is equal to zero. We can see this graphically, where the stationary points of 𝑦=𝑓(π‘₯) have the same π‘₯-value as the π‘₯-intercepts of 𝑦=𝑓′(π‘₯).

This is not the only information we can see from the curve 𝑦=𝑓′(π‘₯). We also know that 𝑓(π‘₯) is increasing when its derivative is positive and decreasing when its derivative is negative. We can also see this information graphically.

This means we can analyze the curve 𝑦=𝑓′(π‘₯) to determine information about the function 𝑓. When 𝑦=𝑓′(π‘₯) has an π‘₯-intercept, 𝑓 must have a stationary point. In other words, the tangent line to 𝑓(π‘₯) at this point is horizontal. This could be a local extremum or it could be a point of inflection.

We know when 𝑓′(π‘₯) is above the π‘₯-axis, we can conclude 𝑓 is increasing. Similarly, when 𝑓′(π‘₯) is below the π‘₯-axis, we can conclude 𝑓 is decreasing.

We can also do this with the second derivative. Recall that the second derivative of a function tells us the concavity of the curve.

Definition: Concavity of a Function

  • We say a function 𝑓(π‘₯) is concave upward on an interval 𝐼 if all of its tangents on this interval lie below the curve. Equivalently, 𝑓′(π‘₯) will be increasing on 𝐼.
  • We say 𝑓(π‘₯) is concave downward on an interval 𝐼 if all of its tangents on this interval lie above the curve. Equivalently, 𝑓′(π‘₯) will be decreasing on 𝐼.

If our function 𝑓 is twice differentiable on an interval, then we can equivalently consider the sign of the second derivative over this interval.

Definition: Concavity of a Twice-Differentiable Function

For a twice-differentiable function 𝑓,

  • if 𝑓′′(π‘₯)>0 on an interval 𝐼, then 𝑓 is concave upward on this interval;
  • if 𝑓′′(π‘₯)<0 on an interval 𝐼, then 𝑓 is concave downward on this interval.

The concavity of a curve may change at values of π‘₯ where its second derivative is equal to zero (or does not exist); these are called points of inflection. We can also see this information graphically.

Let’s start with the curve 𝑦=𝑓′′(π‘₯).

When the curve 𝑦=𝑓′′(π‘₯) is above the π‘₯-axis, the second derivative is positive, so 𝑓 is concave upward. Similarly, when the curve 𝑦=𝑓′′(π‘₯) is below the π‘₯-axis, the second derivative is negative, so 𝑓 is concave downward.

When the second derivative of our function switches from negative to positive, 𝑓 will have a point of inflection and the tangent lines switch from above the curve to below the curve.

We can therefore use the graphs of 𝑦=𝑓′(π‘₯) and 𝑦=𝑓′′(π‘₯) to determine information about the function 𝑓. Let us see an example involving the derivative function.

Example 1: Finding the Monotonicity of a Function given Its Derivative Graph

The graph of the derivative 𝑓′ of a function 𝑓 is shown. On what intervals is 𝑓 increasing or decreasing?

Answer

In this question, we are given the curve of 𝑦=𝑓′(π‘₯) and asked to find the intervals on which 𝑓(π‘₯) is increasing. Usually, we would look at the graph of 𝑓(π‘₯) and look for the portions of our graph where the slope is positive to see where the function is increasing and where the slope is negative to see where the function is decreasing. To answer this question, we need to remember the slope of 𝑓(π‘₯) is given by 𝑓′(π‘₯).

This means we can also see this information in the graph of 𝑦=𝑓′(π‘₯). The derivative, 𝑓′(π‘₯), will be positive when the curve is above the π‘₯-axis and it will be negative when the curve is below the π‘₯-axis.

We can see when π‘₯∈]1,5[ that we have 𝑓′(π‘₯)>0, so the slope of 𝑓(π‘₯) is positive. This means that, for these values of π‘₯, our function 𝑓 must be increasing.

Similarly, when π‘₯∈]0,1[ or π‘₯∈]5,6[, we can see that 𝑓′(π‘₯)<0, so the slope of 𝑓(π‘₯) is negative for these values of π‘₯, which means that 𝑓 is decreasing on these intervals.

Therefore, we were able to show that 𝑓 is increasing on the interval ]1,5[ and decreasing on the intervals ]0,1[ and ]5,6[.

It is worth noting that 𝑓′(1)=0 and 𝑓′(5)=0. Since these values of π‘₯ are the end points of increasing or decreasing intervals, we technically could include these values in our answer.

In fact, some literature always includes endpoints with zero derivative in the intervals where a function increases or decreases. It is personal preference whether or not to include endpoints with zero derivative in increasing or decreasing intervals; we will choose to leave these points out. Also, because our function is not differentiable when π‘₯≀0 and π‘₯β‰₯6, we can just assume that it is not increasing or decreasing for these values either.

Of course, finding increasing and decreasing intervals is not the only thing we use the derivative for. We can also use the second derivative of a function to determine the concavity. Let’s see an example of using the graph of the derivative of a function to find its inflection points.

Example 2: Finding the Inflection Points of a Function from the Graph of Its Derivative

The graph of the first derivative 𝑓′ of a continuous function 𝑓 is shown. State the π‘₯-coordinates of the inflection points of 𝑓.

Answer

In this question, we are tasked with finding the inflection points of the curve 𝑦=𝑓(π‘₯) and to do this we are given a graph of the derivative function 𝑦=𝑓′(π‘₯). We first need to recall that the inflection points are when the curve changes concavity, in other words, where 𝑓′′(π‘₯) changes sign. There are a few different ways to find this from the given graph. The easiest way is to think about the connection between 𝑓′(π‘₯) and 𝑓′′(π‘₯). We know that 𝑓′′(π‘₯) is the derivative of 𝑓′(π‘₯); in other words, 𝑓′′(π‘₯) is the slope of the curve of 𝑦=𝑓′(π‘₯).

We can then ask the following question: what happens when 𝑓′′(π‘₯) changes sign? When 𝑓′′(π‘₯) is positive, the slope of 𝑦=𝑓′(π‘₯) is positive, so it must be increasing. Similarly, when 𝑓′′(π‘₯) is negative, the slope of 𝑦=𝑓′(π‘₯) is negative.

So, we are looking for the points on our graph where the curve switches from increasing to decreasing or vice versa.

We can mark these on our graph; we can see that 𝑓′(π‘₯) is increasing for the following values of π‘₯: π‘₯<2,3<π‘₯<5,π‘₯>7.

Similarly, we can see 𝑓′(π‘₯) is decreasing for the following values of π‘₯: 2<π‘₯<3,5<π‘₯<7.

This means 𝑓′(π‘₯) switches from being increasing to being decreasing when π‘₯=2 and π‘₯=4 and switches from being decreasing to being increasing when π‘₯=3 and π‘₯=7.

Therefore, 𝑓 has inflection points at π‘₯=2, π‘₯=3, π‘₯=5, and π‘₯=7.

We have also seen how to use the first derivative test to determine whether local extrema are maxima or minima. For example, consider the following graph.

We can mark the local extrema as shown, and we can determine their types by considering the slope around these points.

At a local maximum of a differentiable function, the slope changes from positive to negative. At a local minimum of a differentiable function, the slope changes from negative to positive. This gives us a method for checking the type of extremum when we know the derivative function 𝑓′(π‘₯). However, we can use exactly the same reasoning if we are instead given the curve 𝑦=𝑓′(π‘₯). Let us see this in an example.

Example 3: Finding the Values That Give the Local Maximum and Minimum Values of a Function from the Graph of Its First Derivative

The graph of the derivative 𝑓′ of a function 𝑓 is shown. At what values of π‘₯ does 𝑓 have a local maximum or minimum?

Answer

In this question, we are given a graph of 𝑦=𝑓′(π‘₯) and we are asked to determine all the values of π‘₯, where the function 𝑓(π‘₯) has a local extremum. To do this, we need to remember that local extrema must always occur at the critical points of our function, that is, when the derivative is equal to zero or does not exist. In our case, we can see that 𝑓′(π‘₯)=0, when its curve intersects the π‘₯-axis.

So, we have two critical points for our function 𝑓(π‘₯), one at π‘₯=1 and the other at π‘₯=5.

However, we are not done. We still need to check the types of point we have. Since we have a graph of 𝑦=𝑓′(π‘₯), we will do this by using the first derivative test.

Remember, 𝑓′(π‘₯) tells us the slope of the curve 𝑦=𝑓(π‘₯). So, when 𝑓′(π‘₯) is positive, we know the slope of 𝑓(π‘₯) is positive and the same is true in reverse.

By looking at our graph, we can see that

  • when π‘₯<1, 𝑓(π‘₯) has a negative slope;
  • when 1<π‘₯<5, 𝑓(π‘₯) has a positive slope.

The first derivative test then tells us we must have a local minimum at π‘₯=1.

Similarly,

  • when 1<π‘₯<5, 𝑓(π‘₯) has a positive slope;
  • when π‘₯>5, 𝑓(π‘₯) has a negative slope.

This time, the first derivative test then tells that π‘₯=5 is a local maximum for our function 𝑓(π‘₯).

By using the first derivative test graphically, we were able to show that 𝑓 has a local maximum at π‘₯=5 and a local minimum at π‘₯=1.

So far, in all of our examples, we have been given a graph of the function 𝑦=𝑓′(π‘₯). But we can also deduce a lot of information from the graph of the second derivative of a function. Let us see an example of this.

Example 4: Finding the π‘₯-Coordinates of the Inflection Points of a Function from the Graph of Its Second Derivative

Use the given graph of a function 𝑓′′ to find the π‘₯-coordinates of the inflection points of 𝑓.

Answer

We want to find the inflection points of the function 𝑓(π‘₯). Remember, these are points where 𝑓(π‘₯) is continuous and changes concavity, either from concave upward to concave downward or vice versa.

We know all points of inflection occur when 𝑓′′(π‘₯)=0 or when the second derivative does not exist. So, we can see from our diagram this can only happen when π‘₯=1, π‘₯=4, or π‘₯=7.

However, we have only shown that our curve can have inflection points at these values of π‘₯. We still need to check that these are actually inflection points. To do this, we need to check if our curve changes concavity at these values of π‘₯.

We say a curve is concave upward when its second derivative is positive and concave downward when its second derivative is negative. We are given a graph of the curve 𝑦=𝑓′′(π‘₯), so we can see when this is positive or negative by looking at where the curve is above or below the π‘₯-axis.

We can now see that when π‘₯=1, the curve 𝑦=𝑓(π‘₯) will change from concave downward to concave upward. Similarly, when π‘₯=7, the curve 𝑦=𝑓(π‘₯) will change from concave upward to concave downward. So, both of these are inflection points for our curve. However, we can see the concavity does not change from positive to negative or vice versa at π‘₯=4, so this is not an inflection point.

Therefore, we have shown there are two inflection points for the curve 𝑦=𝑓(π‘₯), one when π‘₯=1 and another when π‘₯=7.

Let’s now see an example where we want to determine the intervals where our curve 𝑦=𝑓(π‘₯) is concave upward or concave downward just by using the graph of its derivative.

Example 5: Finding the Concavity of a Function from Its Derivative Graph

The graph of the first derivative 𝑓′ of a function 𝑓 is shown. On what intervals is 𝑓 concave upward or concave downward?

Answer

We want to determine the intervals where the curve 𝑦=𝑓(π‘₯) is concave upward and concave downward; however, instead of being given a graph of this function, we are given the graph of its derivative. So, to answer this question, we will need to start by recalling the connection between the derivative of a function and its concavity.

First, remember, we say a curve is concave upward when its second derivative is positive and concave downward when its second derivative is negative.

This means we need to determine the sign of the second derivative from the graph of the first derivative. To do this, we need to remember that if we differentiate the first derivative, we get the second derivative; in other words, 𝑓′′(π‘₯) is the slope of the curve 𝑦=𝑓′(π‘₯).

Therefore, when the slope of 𝑦=𝑓′(π‘₯) is positive, we must have that the curve 𝑦=𝑓(π‘₯) is concave upward, and when the slope of 𝑦=𝑓′(π‘₯) is negative, we must have that the curve 𝑦=𝑓(π‘₯) is concave downward.

We can mark the intervals where the slope is positive and negative in the graph we were given. We see that 𝑓 is concave upward on ]0,1[, ]2,3[, and ]5,7[ and concave downward on ]1,2[, ]3,5[, and ]7,9[.

Let us finish by recapping some of the main points we have seen.

Key Points

  • We can use the graph of 𝑓′(π‘₯) to find the intervals of increasing and decreasing intervals of 𝑓.
  • We can use the graph of 𝑓′(π‘₯) to find the local extrema of 𝑓 and classify them as maxima or minima.
  • We can use the graph of 𝑓′(π‘₯) or 𝑓′′(π‘₯) to find the intervals on which 𝑓 is concave upward or concave downward.
  • We can use the graph of 𝑓′(π‘₯) or 𝑓′′(π‘₯) to find the inflection points of 𝑓.

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