In previous guides, we learned all about making conclusions regarding the behavior of a function based on the behavior of its derivatives such as whether the function is increasing or decreasing at a point, concave up or concave at a point, and more! While we mostly focused on algebraically determining the behavior of functions, we can also determine information graphically! The key features of the graphs of f, f’, and f’’ are all related to one another. 🔑
Let’s dive into how we can do that!
📈 Connecting a Function, Its First Derivative, and Its Second Derivative
Given the graphs of f, f′, and f′′ or some combination of the three, we can determine information about another much as we did so algebraically. The knowledge you learned in our previous Unit 5 subtopic guides can be carried over to this subtopic—instead of using the equations for f, f′, and f′′, you can look at (one of) their graphs and see where the x-axis is crossed or where the graph is positive or negative, increasing or decreasing, etc, to infer information about the other graphs.
📌 Here is a list of specific guides to revisit:
📉 Trends and Concavity
Here’s a quick summary of what you’ve learned so far in this unit about trends and concavity:
- When a function is increasing, the first derivative will be positive (>0).
- When a function is decreasing, the first derivative will be negative (<0).
- When a function is concave up, the second derivative will be positive (>0) and the first derivative will increase.
- When a function is concave down, the second derivative will be negative (<0) and the first derivative will decrease.
👀 Trends and Concavity Graphically
Let’s apply this information to the following graph of a function, g(x).
Graph of function, g(x), courtesy of Zweig Media
Taking a look at this graph, we can describe g′(x) between each interval:
- From (−∞,−2) and (0.85,2.8), g(x) is decreasing, so g′(x) is negative.
- From (−2,0.85) and (2.8,∞), g(x) is increasing, so g′(x) is positive.
What about g′′(x)? Let’s take a look at the concavity of g(x):
- From (−∞,−0.5) and (1.5,∞), g(x) is concave up. Therefore, g′′(x) is positive and g′(x) is increasing.
- From (−0.5,1.5), g(x) is concave down. Therefore, g′′(x) is negative and g′(x) is decreasing.
🔍 Extrema and Points of Inflection
Based on where the graph of the function changes direction and concavity, we can also interpret maximums, minimums, x-intercepts, and points of inflection of the graphs of the first and second derivatives.
- If f(x) has a relative minimum (the graph changes from decreasing to increasing), then f′(x) will change from negative to positive at that point.
- If f(x) has a relative maximum (the graph changes from increasing to decreasing), then f′(x) will change from positive to negative at that point.
- If f(x) has a point of inflection, changing from concave up to concave down, then f′(x) will have a relative maximum and f′′(x) will change from positive to negative at that point.
- If f(x) has a point of inflection, changing from concave down to concave up, then f′(x) will have a relative minimum and f′′(x) will change from negative to positive at that point.
If we boil this down to two key concepts, realize that:
- All relative extrema of f(x) are x-intercepts of f′(x).
- All points of inflection of f(x) are relative extrema of f′(x).
This may seem like a lot, but once you see it in action, it’ll make more sense! ⬇️
👀 Extrema and POIs Graphically 1
Here’s a relatively easy example! The derivative of the differentiable function f, f′, is graphed.
Graph of the derivative of f with the point (1.5,0) labeled. Image Created with Desmos.
What can we tell about f at the point x=1.5 based on the graph of its derivative f′?
By looking at the graph of f′, we see that f′ crosses the x-axis at the point of interest x=1.5. It is negative before x=1.5 and positive after x=1.5. This means that f is decreasing before the point and increasing after it, indicating that the point x=1.5 is a relative minimum on the graph of f(x).
The justification we used above to determine the answer is essentially just applying the First Derivative Test but in graphical form! Here’s a quick look at f(x) and f′(x) so you can really see their relationship:
Graph of f(x)=x2−3x and its derivative, f′(x)=2x−3. Image Created with Desmos.
👀 Extrema and POIs Graphically 2
Now, let’s take another look at the example before, but focus on relative extrema and points of inflection.
Graph of function, g(x). Image Courtesy of Zweig Media.
You’ll notice the following:
- At x=−2 and x=2.8, g(x) has relative minima. Therefore, g′(x) has x-intercepts at these points and will change from negative to positive.
- At x=0.85, g(x) has a relative maximum. g′(x) has another x-intercept, but the opposite is true: f′(x) will change from positive to negative.
- At x=−0.5, g(x) has a point of inflection, changing from concave up to concave down. This means g′(x) will have a relative maximum and g′′(x) has an x-intercept changing from positive to negative at x=−0.5.
- At x=1.5, g(x) has an inflection point, but it changes from concave down to concave up. Therefore, g′(x) has a relative minimum and g′′(x) has an x-intercept changing from negative to positive at x=1.5.
👀 Extrema and POIs Graphically 3
Before you move on to taking a look at graphs yourself, take a look at the following graph of h(x) and think about:
- What happens to h′(x) at x=−1.3, denoted by the red dotted line?
- What happens to h′(x) at x=−0.667, denoted by the black dotted line?
- What happens to h′′(x) at x=−0.667?
Graph of h(x)=x3+2x2. Image Created with Desmos.
At x=−1.3, h(x) has a relative maximum. This tells us that h′(x) will have an x-intercept at this point and change from positive to negative!
At x=−0.667, h(x) changes from being concave down to concave up. This tells us that h′(x) will have a relative minimum at this point and h′′(x) has an x-intercept, changing from negative to positive.
Take a look at h(x) in blue and h′(x) in green! You can see exactly these trends.
Graph of h(x)=x3+2x2 and h′(x)=3x2+4x. Image Created with Desmos.
And here’s a graph with h′′(x) added as well, denoted in purple.
Graph of h(x)=x3+2x2, h′(x)=3x2+4x, and h′′(x)=6x+4. Image Created with Desmos.
📝Practice Problems
Now it’s time for you to do some practice on your own! These won’t be as tough, they will more generally test your knowledge of these trends.
❓Practice Problems
Question 1:
The second derivative of the differentiable function f, f′′, is graphed.
Graph of the second derivative of f with the point (1,6) labeled. Image Created with Desmos.
Given that f′(1)=0, what can we tell about f at the point x=1 based on the graph of its second derivative f′′?
Question 2:
The second derivative of the differentiable function f, f′′, is graphed.
Graph of the second derivative of f with the point (−1.8,0) labeled. Image Created with Desmos.
What can we tell about f at the point x=−2.4 based on the graph of its second derivative f′′?
✅ Answers and Solutions
Question 1:
Answer: f has a relative minimum at the point x=1.
Solution:
By looking at the graph of f′′, we see that f′′ is positive at the point of interest x=1. This means that f is concave up at the point. Combined with the fact that f′(1)=0, we can apply the Second Derivative Test to conclude that f has a minimum at x=1.5.
Question 2:
Answer: f is concave down at the point x=−2.4.
Solution:
By looking at the graph of f′′, we infer that f′′ is negative at the point of interest x=−2.4. This means that f is concave down at the point.
⭐ Closing
Woah! You made it to the end of this guide. To practice with some of this material, we recommend getting into Desmos and graphing a function, its first derivative, and its second derivative to see the features of each. Good luck! 🍀