What Do You Know?ap Calculus

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Much like the Calculus AB course, the AP Calculus BC course focuses on the unifying themes of calculus. These include derivatives, integrals, limits, approximation, applications and modeling, and sequences and series, as well as basic experience with appropriate methods and applications. Learn AP®︎ Calculus AB for free—everything you need to know about limits, derivatives, and integrals to pass the AP® test. Full curriculum of exercises and videos.

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Section 4-2 : Critical Points

Critical points will show up throughout a majority of this chapter so we first need to define them and work a few examples before getting into the sections that actually use them.

Definition

We say that (x = c) is a critical point of the function (fleft( x right)) if (fleft( c right)) exists and if either of the following are true.

[f'left( c right) = 0hspace{0.5in}{mbox{OR}}hspace{0.5in}f'left( c right),{mbox{doesn't exist}}]

Note that we require that (fleft( c right)) exists in order for (x = c) to actually be a critical point. This is an important, and often overlooked, point. What this is really saying is that all critical points must be in the domain of the function. If a point is not in the domain of the function then it is not a critical point.

Note as well that, at this point, we only work with real numbers and so any complex numbers that might arise in finding critical points (and they will arise on occasion) will be ignored. There are portions of calculus that work a little differently when working with complex numbers and so in a first calculus class such as this we ignore complex numbers and only work with real numbers. Calculus with complex numbers is beyond the scope of this course and is usually taught in higher level mathematics courses.

The main point of this section is to work some examples finding critical points. So, let’s work some examples.

Example 1 Determine all the critical points for the function. [fleft( x right) = 6{x^5} + 33{x^4} - 30{x^3} + 100] Show Solution

We first need the derivative of the function in order to find the critical points and so let’s get that and notice that we’ll factor it as much as possible to make our life easier when we go to find the critical points.

[begin{align*}f'left( x right) & = 30{x^4} + 132{x^3} - 90{x^2} & = 6{x^2}left( {5{x^2} + 22x - 15} right) & = 6{x^2}left( {5x - 3} right)left( {x + 5} right)end{align*}]

Now, our derivative is a polynomial and so will exist everywhere. Therefore, the only critical points will be those values of (x) which make the derivative zero. So, we must solve.

[6{x^2}left( {5x - 3} right)left( {x + 5} right) = 0]

Because this is the factored form of the derivative it’s pretty easy to identify the three critical points. They are,

[x = - 5,x = 0,x = frac{3}{5}]

Polynomials are usually fairly simple functions to find critical points for provided the degree doesn’t get so large that we have trouble finding the roots of the derivative.

Most of the more “interesting” functions for finding critical points aren’t polynomials however. So let’s take a look at some functions that require a little more effort on our part.

Example 2 Determine all the critical points for the function. [gleft( t right) = sqrt[3]{{{t^2}}}left( {2t - 1} right)] Show Solution

To find the derivative it’s probably easiest to do a little simplification before we actually differentiate. Let’s multiply the root through the parenthesis and simplify as much as possible. This will allow us to avoid using the product rule when taking the derivative.

[gleft( t right) = {t^{frac{2}{3}}}left( {2t - 1} right) = 2{t^{frac{5}{3}}} - {t^{frac{2}{3}}}]

Now differentiate.

[g'left( t right) = frac{{10}}{3}{t^{frac{2}{3}}} - frac{2}{3}{t^{ - frac{1}{3}}} = frac{{10{t^{frac{2}{3}}}}}{3} - frac{2}{{3{t^{frac{1}{3}}}}}]

We will need to be careful with this problem. When faced with a negative exponent it is often best to eliminate the minus sign in the exponent as we did above. This isn’t really required but it can make our life easier on occasion if we do that.

Notice as well that eliminating the negative exponent in the second term allows us to correctly identify why (t = 0) is a critical point for this function. Once we move the second term to the denominator we can clearly see that the derivative doesn’t exist at (t = 0) and so this will be a critical point. If you don’t get rid of the negative exponent in the second term many people will incorrectly state that (t = 0) is a critical point because the derivative is zero at (t = 0). While this may seem like a silly point, after all in each case (t = 0) is identified as a critical point, it is sometimes important to know why a point is a critical point. In fact, in a couple of sections we’ll see a fact that only works for critical points in which the derivative is zero.

So, we’ve found one critical point (where the derivative doesn’t exist), but we now need to determine where the derivative is zero (provided it is of course…). To help with this it’s usually best to combine the two terms into a single rational expression. So, getting a common denominator and combining gives us,

[g'left( t right) = frac{{10t - 2}}{{3{t^{frac{1}{3}}}}}]

Notice that we still have (t = 0) as a critical point. Doing this kind of combining should never lose critical points, it’s only being done to help us find them. As we can see it’s now become much easier to quickly determine where the derivative will be zero. Recall that a rational expression will only be zero if its numerator is zero (and provided the denominator isn’t also zero at that point of course).

So, in this case we can see that the numerator will be zero if (t = frac{1}{5}) and so there are two critical points for this function.

[t = 0hspace{0.5in},{mbox{and}}hspace{0.5in}t = frac{1}{5}] Example 3 Determine all the critical points for the function. [Rleft( w right) = frac{{{w^2} + 1}}{{{w^2} - w - 6}}] Show Solution

We’ll leave it to you to verify that using the quotient rule, along with some simplification, we get that the derivative is,

[R'left( w right) = frac{{ - {w^2} - 14w + 1}}{{{{left( {{w^2} - w - 6} right)}^2}}} = - frac{{{w^2} + 14w - 1}}{{{{left( {{w^2} - w - 6} right)}^2}}}]

Notice that we factored a “-1” out of the numerator to help a little with finding the critical points. This negative out in front will not affect the derivative whether or not the derivative is zero or not exist but will make our work a little easier.

Now, we have two issues to deal with. First the derivative will not exist if there is division by zero in the denominator. So we need to solve,

[{w^2} - w - 6 = left( {w - 3} right)left( {w + 2} right) = 0]

We didn’t bother squaring this since if this is zero, then zero squared is still zero and if it isn’t zero then squaring it won’t make it zero.

So, we can see from this that the derivative will not exist at (w = 3) and (w = - 2). However, these are NOT critical points since the function will also not exist at these points. Recall that in order for a point to be a critical point the function must actually exist at that point.

At this point we need to be careful. The numerator doesn’t factor, but that doesn’t mean that there aren’t any critical points where the derivative is zero. We can use the quadratic formula on the numerator to determine if the fraction as a whole is ever zero.

[w = frac{{ - 14 pm sqrt {{{left( {14} right)}^2} - 4left( 1 right)left( { - 1} right)} }}{{2left( 1 right)}} = frac{{ - 14 pm sqrt {200} }}{2} = frac{{ - 14 pm 10sqrt 2 }}{2} = - 7 pm 5sqrt 2 ]

So, we get two critical points. Also, these are not “nice” integers or fractions. This will happen on occasion. Don’t get too locked into answers always being “nice”. Often they aren’t.

Note as well that we only use real numbers for critical points. So, if upon solving the quadratic in the numerator, we had gotten complex number these would not have been considered critical points.

Summarizing, we have two critical points. They are,

[w = - 7 + 5sqrt 2 ,w = - 7 - 5sqrt 2 ]

Again, remember that while the derivative doesn’t exist at (w = 3) and (w = - 2) neither does the function and so these two points are not critical points for this function.

In the previous example we had to use the quadratic formula to determine some potential critical points. We know that sometimes we will get complex numbers out of the quadratic formula. Just remember that, as mentioned at the start of this section, when that happens we will ignore the complex numbers that arise.

So far all the examples have not had any trig functions, exponential functions, etc. in them. We shouldn’t expect that to always be the case. So, let’s take a look at some examples that don’t just involve powers of (x).

Example 4 Determine all the critical points for the function. [y = 6x - 4cos left( {3x} right)] Show Solution

First get the derivative and don’t forget to use the chain rule on the second term.

[y' = 6 + 12sin left( {3x} right)]

Now, this will exist everywhere and so there won’t be any critical points for which the derivative doesn’t exist. The only critical points will come from points that make the derivative zero. We will need to solve,

[begin{align*}6 + 12sin left( {3x} right) & = 0 sin left( {3x} right) & = - frac{1}{2}end{align*}]

Solving this equation gives the following.

[begin{align*}3x & = 3.6652 + 2pi n,hspace{0.25in}n = 0, pm 1, pm 2, ldots 3x & = 5.7596 + 2pi n,hspace{0.25in}n = 0, pm 1, pm 2, ldots end{align*}]

Don’t forget the (2 pi n) on these! There will be problems down the road in which we will miss solutions without this! Also make sure that it gets put on at this stage! Now divide by 3 to get all the critical points for this function.

[begin{align*}x &= 1.2217 + frac{{2pi n}}{3},hspace{0.5in}n = 0, pm 1, pm 2, ldots x &= 1.9199 + frac{{2pi n}}{3},hspace{0.5in}n = 0, pm 1, pm 2, ldots end{align*}]

Notice that in the previous example we got an infinite number of critical points. That will happen on occasion so don’t worry about it when it happens.

Example 5 Determine all the critical points for the function. [hleft( t right) = 10t{{bf{e}}^{3 - {t^2}}}] Show Solution

Here’s the derivative for this function.

[h'left( t right) = 10{{bf{e}}^{3 - {t^2}}} + 10t{{bf{e}}^{3 - {t^2}}}left( { - 2t} right) = 10{{bf{e}}^{3 - {t^2}}} - 20{t^2}{{bf{e}}^{3 - {t^2}}}]

Now, this looks unpleasant, however with a little factoring we can clean things up a little as follows,

[h'left( t right) = 10{{bf{e}}^{3 - {t^2}}}left( {1 - 2{t^2}} right)]

This function will exist everywhere, so no critical points will come from the derivative not existing. Determining where this is zero is easier than it looks. We know that exponentials are never zero and so the only way the derivative will be zero is if,

[begin{align*}1 - 2{t^2} & = 0 1 & = 2{t^2} frac{1}{2} & = {t^2}end{align*}]

We will have two critical points for this function.

[t = pm frac{1}{{sqrt 2 }}] Example 6 Determine all the critical points for the function. [fleft( x right) = {x^2}ln left( {3x} right) + 6] Show Solution

Before getting the derivative let’s notice that since we can’t take the log of a negative number or zero we will only be able to look at (x > 0).

The derivative is then,

[begin{align*}f'left( x right) & = 2xln left( {3x} right) + {x^2}left( {frac{3}{{3x}}} right) & = 2xln left( {3x} right) + x & = xleft( {2ln left( {3x} right) + 1} right)end{align*}]

Now, this derivative will not exist if (x) is a negative number or if (x = 0), but then again neither will the function and so these are not critical points. Remember that the function will only exist if (x > 0) and nicely enough the derivative will also only exist if (x > 0) and so the only thing we need to worry about is where the derivative is zero.

First note that, despite appearances, the derivative will not be zero for (x = 0). As noted above the derivative doesn’t exist at (x = 0) because of the natural logarithm and so the derivative can’t be zero there!

So, the derivative will only be zero if,

[begin{align*}2ln left( {3x} right) + 1 & = 0 ln left( {3x} right) & = - frac{1}{2}end{align*}]

Recall that we can solve this by exponentiating both sides.

[begin{align*}{{bf{e}}^{ln left( {3x} right)}} & = {{bf{e}}^{ - frac{1}{2}}} 3x & = {{bf{e}}^{ - frac{1}{2}}} x & = frac{1}{3}{{bf{e}}^{ - frac{1}{2}}} = frac{1}{{3sqrt {bf{e}} }}end{align*}]

There is a single critical point for this function.

Let’s work one more problem to make a point.

Example 7 Determine all the critical points for the function. [fleft( x right) = x{{bf{e}}^{{x^2}}}] Show Solution

Note that this function is not much different from the function used in Example 5. In this case the derivative is,

[f'left( x right) = {{bf{e}}^{{x^2}}} + x{{bf{e}}^{{x^2}}}left( {2x} right) = {{bf{e}}^{{x^2}}}left( {1 + 2{x^2}} right)]

This function will never be zero for any real value of (x). The exponential is never zero of course and the polynomial will only be zero if (x) is complex and recall that we only want real values of (x) for critical points.

Therefore, this function will not have any critical points.

It is important to note that not all functions will have critical points! In this course most of the functions that we will be looking at do have critical points. That is only because those problems make for more interesting examples. Do not let this fact lead you to always expect that a function will have critical points. Sometimes they don’t as this final example has shown.

I have a love/hate relationship with calculus: it demonstrates the beauty of math and the agony of math education.

Calculus relates topics in an elegant, brain-bending manner. My closest analogy is Darwin’s Theory of Evolution: once understood, you start seeing Nature in terms of survival. You understand why drugs lead to resistant germs (survival of the fittest). You know why sugar and fat taste sweet (encourage consumption of high-calorie foods in times of scarcity). It all fits together.

Calculus is similarly enlightening. Don’t these formulas seem related in some way?

They are. But most of us learn these formulas independently. Calculus lets us start with $text{circumference} = 2 pi r$ and figure out the others — the Greeks would have appreciated this.

Unfortunately, calculus can epitomize what’s wrong with math education. Most lessons feature contrived examples, arcane proofs, and memorization that body slam our intuition & enthusiasm.

It really shouldn’t be this way.

Math, art, and ideas

I’ve learned something from school: Math isn’t the hard part of math; motivation is. Specifically, staying encouraged despite

• Teachers focused more on publishing/perishing than teaching
• Self-fulfilling prophecies that math is difficult, boring, unpopular or “not your subject”
• Textbooks and curriculums more concerned with profits and test results than insight

‘A Mathematician’s Lament’ [pdf] is an excellent essay on this issue that resonatedwithmany people:

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“…if I had to design a mechanism for the express purpose of destroying a child’s natural curiosity and love of pattern-making, I couldn’t possibly do as good a job as is currently being done — I simply wouldn’t have the imagination to come up with the kind of senseless, soul-crushing ideas that constitute contemporary mathematics education.”

Imagine teaching art like this: Kids, no fingerpainting in kindergarten. Instead, let’s study paint chemistry, the physics of light, and the anatomy of the eye. After 12 years of this, if the kids (now teenagers) don’t hate art already, they may begin to start coloring on their own. After all, they have the “rigorous, testable” fundamentals to start appreciating art. Right?

Poetry is similar. Imagine studying this quote (formula):

“This above all else: to thine own self be true, and it must follow, as night follows day, thou canst not then be false to any man.” —William Shakespeare, Hamlet

It’s an elegant way of saying “be yourself” (and if that means writing irreverently about math, so be it). But if this were math class, we’d be counting the syllables, analyzing the iambic pentameter, and mapping out the subject, verb and object.

Math and poetry are fingers pointing at the moon. Don’t confuse the finger for the moon. Formulas are a means to an end, a way to express a mathematical truth.

We’ve forgotten that math is about ideas, not robotically manipulating the formulas that express them.

Ok bub, what’s your great idea?

Feisty, are we? Well, here’s what I won’t do: recreate the existing textbooks. If you need answers right away for that big test, there’s plenty of websites, class videos and 20-minute sprints to help you out.

Instead, let’s share the core insights of calculus. Equations aren’t enough — I want the “aha!” moments that make everything click.

Formal mathematical language is one just one way to communicate. Diagrams, animations, and just plain talkin’ can often provide more insight than a page full of proofs.

But calculus is hard!

I think anyone can appreciate the core ideas of calculus. We don’t need to be writers to enjoy Shakespeare.

It’s within your reach if you know algebra and have a general interest in math. Not long ago, reading and writing were the work of trained scribes. Yet today that can be handled by a 10-year old. Why?

Because we expect it. Expectations play a huge part in what’s possible. So expect that calculus is just another subject. Some people get into the nitty-gritty (the writers/mathematicians). But the rest of us can still admire what’s happening, and expand our brain along the way.

It’s about how far you want to go. I’d love for everyone to understand the core concepts of calculus and say “whoa”.

So what’s calculus about?

Some define calculus as “the branch of mathematics that deals with limits and the differentiation and integration of functions of one or more variables”. It’s correct, but not helpful for beginners.

Here’s my take: Calculus does to algebra what algebra did to arithmetic.

• Arithmetic is about manipulating numbers (addition, multiplication, etc.).

• Algebra finds patterns between numbers: $a^2 + b^2 = c^2$ is a famous relationship, describing the sides of a right triangle. Algebra finds entire sets of numbers — if you know a and b, you can find c.

• Calculus finds patterns between equations: you can see how one equation ($text{circumference} = 2 pi r$) relates to a similar one ($text{area} = pi r^2$).

Using calculus, we can ask all sorts of questions:

• How does an equation grow and shrink? Accumulate over time?
• When does it reach its highest/lowest point?
• How do we use variables that are constantly changing? (Heat, motion, populations, …).
• And much, much more!

Algebra & calculus are a problem-solving duo: calculus finds new equations, and algebra solves them. Like evolution, calculus expands your understanding of how Nature works.

An Example, Please

Let’s walk the walk. Suppose we know the equation for circumference ($2 pi r$) and want to find area. What to do?

Realize that a filled-in disc is like a set of Russian dolls.

Here are two ways to draw a disc:

• Make a circle and fill it in
• Draw a bunch of rings with a thick marker

The amount of “space” (area) should be the same in each case, right? And how much space does a ring use?

Well, the very largest ring has radius “r” and a circumference $2 pi r$. As the rings get smaller their circumference shrinks, but it keeps the pattern of $2 pi cdot text{current radius}$. The final ring is more like a pinpoint, with no circumference at all.

Now here’s where things get funky. Let’s unroll those rings and line them up. What happens?

• We get a bunch of lines, making a jagged triangle. But if we take thinner rings, that triangle becomes less jagged (more on this in future articles).
• One side has the smallest ring (0) and the other side has the largest ring ($2 pi r$)
• We have rings going from radius 0 to up to “r”. For each possible radius (0 to r), we just place the unrolled ring at that location.
• The total area of the “ring triangle” = $frac{1}{2} text{ base} cdot text{height} = frac{1}{2} (r) (2 pi r) = pi r^2$, which is the formula for area!

Yowza! The combined area of the rings = the area of the triangle = area of circle!

(Image from Wikipedia)

This was a quick example, but did you catch the key idea? We took a disc, split it up, and put the segments together in a different way. Calculus showed us that a disc and ring are intimately related: a disc is really just a bunch of rings.

This is a recurring theme in calculus: Big things are made from little things. And sometimes the little things are easier to work with.

A note on examples

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Many calculus examples are based on physics. That’s great, but it can be hard to relate: honestly, how often do you know the equation for velocity for an object? Less than once a week, if that.

I prefer starting with physical, visual examples because it’s how our minds work. That ring/circle thing we made? You could build it out of several pipe cleaners, separate them, and straighten them into a crude triangle to see if the math really works. That’s just not happening with your velocity equation.

A note on rigor (for the math geeks)

I can feel the math pedants firing up their keyboards. Just a few words on “rigor”.

Did you know we don’t learn calculus the way Newton and Leibniz discovered it? They used intuitive ideas of “fluxions” and “infinitesimals” which were replaced with limits because “Sure, it works in practice. But does it work in theory?”.

We’ve created complex mechanical constructs to “rigorously” prove calculus, but have lost our intuition in the process.

We’re looking at the sweetness of sugar from the level of brain-chemistry, instead of recognizing it as Nature’s way of saying “This has lots of energy. Eat it.”

I don’t want to (and can’t) teach an analysis course or train researchers. Would it be so bad if everyone understood calculus to the “non-rigorous” level that Newton did? That it changed how they saw the world, as it did for him?

A premature focus on rigor dissuades students and makes math hard to learn. Case in point: e is technically defined by a limit, but the intuition of growth is how it was discovered. The natural log can be seen as an integral, or the time needed to grow. Which explanations help beginners more?

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Let’s fingerpaint a bit, and get into the chemistry along the way. Happy math.

(PS: A kind reader has created an animated powerpoint slideshow that helps present this idea more visually (best viewed in PowerPoint, due to the animations). Thanks!)

Note: I’ve made an entire intuition-first calculus series in the style of this article:

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