Finding Critical Points in Calculus: Function & Graph

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  • 0:05 What are Critical Points?
  • 3:31 Example with Graph
  • 5:28 Lesson Summary
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Lesson Transcript
Jason Furney

Jason has taught both College and High School Mathematics and holds a Master's Degree in Math Education.

Expert Contributor
Kathryn Boddie

Kathryn earned her Ph.D. in Mathematics from UW-Milwaukee in 2019. She has over 10 years of teaching experience at high school and university level.

This lesson develops the understanding of what a critical point is and how they are found. It explores the definition and discovery of critical points using functions and graphs as well as possible uses for them in the everyday world.

What Are Critical Points?

Critical points are key in calculus to find maximum and minimum values of graphs.

Let's say you bought a new dog, and went down to the local hardware store and bought a brand new fence for your yard, but alas, it doesn't come assembled. Of course, this means that you get to fence in whatever size lot you want with restrictions of how much fence you have. Wouldn't you want to maximize the amount of space your dog had to run? Critical points can tell you the exact dimensions of your fenced-in yard that will give you the maximum area!

Critical points in calculus have other uses, too. For example, they could tell you the lowest or highest point of a suspension bridge (assuming you can plot the bridge on a coordinate plane). Now we know what they can do, but how do we find them? First, let's officially define what they are.

Definition of a Critical Point

Let f be defined at b. If f(b) = 0 or if 'f' is not differentiable at b, then b is a critical number of f. If this critical number has a corresponding y value on the function f, then a critical point exists at (b, y).

What exactly does this mean? Well, f just represents some function, and b represents the point or the number we're looking for. The second part of the definition tells us that we can set the derivative of our function equal to zero and solve for x to get the critical number! The third part says that critical numbers may also show up at values in which the derivative does not exist. We'll look at an example of this a bit later. Lastly, if the critical number can be plugged back into the original function, the x and y values we get will be our critical points.

Finding Critical Points

Now we're going to look at a graph, point out some critical points, and try to find why we set the derivative equal to zero.

The red dots on the graph represent the critical points of that particular function, f(x). It's here where you should start asking yourself a few questions:

  • Is there something similar about the locations of both critical points? You should look for visual similarities.
  • How does this compare to the definition from above?

If you understand the answers to these two questions, then you can understand how we find critical points.

Notice how both critical points tend to appear on a hump or curve of the graph. More specifically, they are located at the very top or bottom of these humps. Mathematically speaking, the slope changes from positive to negative (or vice versa) at these points. It's why they are so critical!

To understand how number one relates to the defection of a critical point, we have to remember what exactly a derivative tells us. The derivative of a function, f(x), gives us a new function f(x) that represents the slopes of the tangent lines at every specific point in f(x). So why do we set those derivatives equal to 0 to find critical points? Take a look at the following graph that shows different tangent lines to f(x):

The green tangent lines run through our critical points. What's the difference between those and the blue ones? For one thing, they have the same slope, whereas the blue tangent lines all have different slopes. For another thing, that slope is always one very specific number. Who remembers the slope of a horizontal line? That's right! The slope of every tangent line that passes through a critical point is always 0!

Example with Graph

Find the critical points of the following:


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Additional Activities

Finding Critical Points to Maximize a Fence in Yard

Now that we understand how to find critical points, let's use our knowledge to answer the question posed at the beginning of the video lesson. Suppose you just got a new dog and you plan to build a fence in your yard for the dog to run in. You have to assemble the fence by yourself. You purchased 50 feet of fence and wish to enclose a rectangular area for your dog. You only need fence along 3 sides of the rectangle - your house encloses the fourth side. What are the dimensions you should enclose to maximize the area for your dog to run in?

We know a function has a local minimum or local maximum at its critical points - so let's find a function to represent the area that is being enclosed, find the critical points of the function, and make sure we find a maximum!

Steps to Maximizing the Area

Perform these steps on your own and check your work with the guide below the steps.

1. Find an equation representing the area in this situation. To do this, write an equation for area and an equation representing the length of the fence. Make sure all information is used and the area equation has only one variable before moving to step 2.

2. Take the derivative of the area equation.

3. Find the critical numbers.

4. Verify that the area function is maximized at a critical number.

5. Find the other variable to find the other dimension of the rectangle.

Setting up the Equations

We know that the area of a rectangle is given by A = xy where x is the length and y is the width. But we need more information than this - how can we use the 50 feet of fence in another equation? We are adding up three sides of a rectangle, and it must use all 500 feet of fence, so we have 50 = 2x + y where x is the length of the parallel sides and y is the length of the side opposite the house.

To write the area equation with just one variable, take the equation about the length of the fence and solve it for y. Then substitute this information into the area equation.

50 = 2x + y

50 - 2x = y

Substituting in, we have

A = xy

A = x(50 - 2x)

A = 50x - 2x2

Finding the Critical Numbers

Take the derivative of the area function and find where it is zero or undefined.

A = 50x - 2x2

A' = 50 - 4x

0 = 50 - 4x

4x = 50

x = 12.5

We have a critical value at x = 12.5. To verify that this information will give us a maximum, let's see what happens if we substitute a value smaller than 12.5 and a value larger than 12.5 into the derivative. We know a maximum will occur if the function changes from increasing to decreasing - that is, if the derivative changes from positive to negative.

A'(10) = 50 - 4(10) = 10 > 0

A'(15) = 50 - 4(15) = -10 < 0

Since the derivative changes from positive to negative when x = 12.5, we have a maximum.

Solving the Final Answer

We know that one dimension is 12.5 feet. To find the other, substitute this back in to the equation about the fence length.

50 = 2x + y

50 = 2(12.5) + y

50 = 25 + y

25 = y

The dimensions that should be enclosed are 12.5 * 25 feet (with the longer side opposite the house) for a maximum area of A = 12.5 * 25 = 312.5 square feet.

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