Essential Knowledge 2.2A1

Essential Knowledge 2.2A1 Students will know that first and second derivatives of a function can provide information about the function and its graph including intervals of increase or decrease, local (relative) and global (absolute) extrema, intervals of upward or downward concavity, and points of inflection.

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Introduction

Thus far in your education, you've only really learned one way to graph a complicated equation. Namely, you make a table of x-values, use substitution to solve for the corresponding y-values, and graph the ordered pairs. This can take a long time, and has huge chances to fail. For example, what if you didn't graph enough points near a peak, and now you don't know where the peak actually is, and it's much shorter than it should be.

Calculus lets you take a new approach to graphing complex equations. I think it's brilliant and I use this method every chance I get.

We'll learn the ways to use derivatives to help create better graphs here.

Here's the idea. Ask yourself this: when sketching the graph of an unfamiliar curve, what information would help the most? While there is technically no correct answer to this question, derivatives give two very useful pieces of information.

  • The derivative allows you to find the relative maxima and minima. These are also known as the critical points. (Though not all critical points are relative maxima or minima. In other words, all max's and min's are critical points, but not all critical points are max's or min's.)

  • The second derivative allows you to find the points of inflection

Some vocabulary

  • Relative maximum (plural: relative maxima) a point where a curve "peaks". A peak is a point where the graph was rising, then began to fall.

  • Relative minimum (plural: relative minima) a point where a curve "troughs". A trough is a point where the graph was falling, then began to rise.

Example

Use derivatives to estimate the graph of f(x) = 3x5 - 5x3 + 3.

Critical Points

First thing we want to do is find where the peaks and troughs are on this graph.

First, go visit Essential Knowledge 2.1B1 and take a look at the GeoGebra applet. At every peak and trough, what is the slope of the tangent line? As you can see, the tangent line always goes flat at these parts of the curve, meaning the tangent line has a slope of zero at all peaks and troughs

So, we know the slope of the tangent line to a function equals the derivative of the function. If we want to know where the slope is zero, let's take the derivative of the function and see when it is zero.

f(x) = 3x5 - 5x3 + 3

f '(x) = 15x4 - 15x2

Let f '(x) = 0

0 = 15x4 - 15x2

Now if we want to find where points of zero slope occur, we simply need to solve for x. This almost always requires factoring.

0 = 15x4 - 15x2

0=15x2(x2 - 1)

0=15x2(x + 1)(x - 1)

x = -1, 0, 1

In other words, there might be either a peak or trough when x = -1, x = 0, and x = 1. Substituting, we can find the y-values and thus plot those points.

f(-1) = 3(-1)5 - 5(-1)3 + 3

f(-1) = -3 + 5 + 3

f(-1) = 5

f(0) = 3(0)5 - 5(0)3 + 3

f(0) = 3

f(1) = 3x5 - 5x3 + 3

f(1) = 3(1)5 - 5(1)3 + 3

f(1) = 3 - 5 + 3

f(1) = 1

Thus the re are critical points at (-1, 5), (0, 3), and (1, 1). Let's plot these important points.

We want to know which are relative maxima and which are relative minima. Remember, a peak happens when the curve was rising then began to fall. How do you know if a curve is rising or falling. Simple: the slope of the tangent line is positive when a curve is rising, and negative when a curve is falling. So, we just need to find derivatives before and after these critical points.

To see if (-1, 5) is a peak, trough, or neither, let's find the slope at x = -1.5 and x = 0.5.

Let x = -1.5

f'(-1.5) = 15(-1.5)4 - 15(-1.5)2

f'(-1.5) = 15(5.0625) - 15(2.25)

f'(-1.5) = 75.9375 - 33.75

f'(-1.5) = 42.1875

Let x = -0.5

f'(-0.5) = 15(-0.5)4 - 15(-0.5)2

f'(-0.5) = 15(0.0625) - 15(0.25)

f'(-0.5) = 0.9375 - 3.75

f'(-0.5) = -2.8125

We can see that the curve was rising, had a slope of 0, then began falling. Thus, (-1, 5) must have been a max.

Now let's do the same for (0, 3). We've already got a slope to the left (which is that f'(-0.5) is negative). So now let's do one to the right, such as x = 0.5

Let x = 0.5

f'(0.5) = 15(0.5)4 - 15(0.5)2

f'(0.5) = 15(0.0625) - 15(0.25)

f'(0.5) = 0.9375 - 3.75

f'(0.5) = -2.8125

Interesting. The curve was falling, then had a slope of zero, then continued to fall. This point, even though it is a critical point, is not a peak or a trough. False alarm.

Lastly, let's investigate the critical point at x = 1. We already have a value to the left (namely x = 0.5), so let's do x = 1.5.

Let x = 1.5

f'(1.5) = 15(1.5)4 - 15(1.5)2

f'(1.5) = 15(5.0625) - 15(2.25)

f'(1.5) = 75.9375 - 33.75

f'(1.5) = 42.1875

We can see that the curve was falling, had a slope of 0, then began rising. Thus, (1, 1) must have been a min.

Points of Inflection

Now we want to find the points of inflection. A point of inflection marks a location where a graph went from being concave up to concave down, or visa versa. This is a bit harder to explain than critical points. Your best bet is to have several different ways of thinking about concavity, and eventually all of them will make sense

  • A curve is concave up if the slope is increasing. A curve is concave down if the slope is decreasing.

    • This is a precise definition, but takes some practice to wield.

  • A curve is concave up if it looks like a upward facing parabola. A curve is concave down if it looks like a downward facing parabola.

    • This is not a precise definition, because parabolas aren't the only shapes that change slope.

    • Nonetheless, this is a great way to visualize concavity

Here's a familiar graph that you should link in your mind to "concave up"

At all times along the curve, when going from left to right the slope is becoming more positive. At the beginning the slope is very negative. As the curve approaches x = 0, the slope increases slowly to 0. At x = 0 the slope is 0. Then, after x = 0, the slope is slightly positive. Then it becomes more positive.

Of course for concave down, you should train yourself to think of these.

Now how will we find where these points occur. Let's think about the definition. We need a way to determine if the slope is increasing or decreasing. Well, if we want to know if a function is increasing or decreasing, we know that is the slope of the tangent line, and therefore we take the derivative of the function. So if the slope is the derivative, and we want to know if a slope is increasing or decreasing, we need the slope of the derivative. In other words, the derivative of the derivative. In other words, the second derivative.

f(x) = 3x5 - 5x3 + 3

f '(x) = 15x4 - 15x2

f ''(x) = 60x3 - 30x

A point of inflection marks where the concavity changes. In order for a concavity to go from concave up (positive) to concave down (negative) or visa versa, it must equal zero at the transition. Therefore points of inflection occur when the second derivative equals zero. So let's find these locations.

Let f''(x) = 0

Let's find the corresponding y-values so we can plot these points.

f(0) = 3(0)5 - 5(0)3 + 3

f(0) = 3

Thus, the graph will have inflection points at (-√(1/2), 4.2375), (0, 3), and (√(1/2), 1.7625). Let's add these to our graph.

The last thing we need to know is where the curve is concave up and concave down. We can figure out most of this logically. Since there is a peak at x = -1, we know that that region is concave down, because peaks can only occur in regions that are concave down. The curve will need to stay concave down until it reaches an inflection point. This happens at x=-√(1/2). So let's see what happens after this inflection point. For example, let's try x = -0.5

Let x = -0.5

f ''(x) = 60(-0.5)3 - 30(-0.5) = - 7.5 + 15 = 7.5

We get a positive value here, so the concavity must be upward between x=-√(1/2) and x = 0. Using the same logic, we need to check between x = 0 and x=√(1/2). For example, let's try x = 0.5

f ''(x) = 60(0.5)3 - 30(0.5) = 7.5 - 15 = -7.5

We get a negative value here, so the concavity must be downward between x = 0 and x=√(1/2)

Now we know enough to try to sketch this curve. Let's summarize what we've learned in a table to prepare ourselves.

  • So, I'll just follow the story of my table. Looks like I need to start off with a positive slope until I peak at x = -1. At x = -1, the slope must be zero.

  • Then I continue with a concave down curve until x=-√(1/2).

  • Then I continue downward until x = 0, but now with a concave up slope. At x = 0, the slope must be zero

  • Then I continue downward until x=√(1/2), but now with a concave down slope.

  • Then I continue downward until x = 1, but now with a concave up slope. At x = 1, the slope must be zero.

  • From this point on, there is just a positive slope

Behold my creation. Notice that concave down parts are red and concave up parts are blue, just to help you learn to identify them visually

Now let's add in the actual graph to see how I did.

Your first impression may be that I was way off. But when you think about how you normally use graphs, and which parts tend to be important, you'll start to realize that this is actually a very good approximation. Especially considering that I only found the coordinates of 5 points! This use of calculus doesn't guarantee a perfect graph, but it does capture the important parts of the graph and the general shape. For the time I put in, this is a great approximation.

You may have noticed we could have just done concavity to see if critical points were peaks or troughs. Feel free to do that instead, it's actually faster.