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Determining the End Behavior of a Function

How do you determine the end behavior of a function?  And, what does this mean?

When looking at a graph, the “end behavior” is referring to what is happening all the way to the far left of the graph and all the way to the far right of the graph.  Your goal is to analyze the y-value (height or function value) of the function when x is really large and negative, and then again when x is really large and positive.  What is the pattern on each end?  What is the “end behavior”?

Notationally, we are thinking:

1. As x → -∞, y → ?
2. As x → +∞, y → ?

OK, so let’s try this on a polynomial example:

Q:  What is the end behavior of the function y=5x3+7x2-2x-1

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Limits vs Values

Look at the function f(x) in orange below:

We are going to answer 4 questions about this graph.  They are all related to each other, but different questions.  Seeing the difference will help us sort out the difference between a function value and a limit.

Q1:  Find f(1)

Q2:  Find  limx→1 f(x)

Q3:  Find  limx→1+ f(x)

Q4:  Find  limx→1 f(x)

OK….. Try to answer these questions with what you know… Then continue reading to see the answers and explanations!

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Basic Concept of a Limit

Here is a brief example on the concept of a limit:

Look at the function f(x) in orange below:

Question 1.  Find f(3)

Explanation of question 1: Find the value of the function when you plug in 3.  What is the height of the function at the exact moment when x=3?

Answer 1:  The function is undefined at x=3.  There is a hole when x=3.

So, f(3) is undefined.

Question 2:  Find limx→3f(x)

Explanation of question 2:  We are being asked to find what the function is doing around (but not at) 3.  What is happening to the path of the function on either side of 3?

In order to find limx→3f(x), we must confirm that limx→3+  f(x) and limx→3–  f(x) both exist and are equal to each other.

So, let’s find limx→3+  f(x).  What is happening to the function values as you approach x=3 from the right-hand side?  Literally run your finger along as if x=4, then x=3.5, then x=3.1.  What value is the function getting closer to?

The function is approaching a height of 4.

Let’s find limx→3–  f(x).  What is happening to the function values as you approach x=3 from the left-hand side?  Literally run your finger along as if x=1, then x=2, then x=2.9.  What value is the function getting closer to?

The function is also approaching a height of 4.

So:

limx→3+  f(x) = 4

limx→3–  f(x) = 4

Since, the left-handed limit at 3 and right-handed limit at 3 exist and are equal, this gives:

limx→3f(x) = 4.

So, to summarize, here are 4 different things we found.  They are related, but not necessarily the same:

f(3) is undefined

limx→3+  f(x) = 4

limx→3–  f(x) = 4

limx→3f(x) = 4

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Limit Example

Q:  What is limx→2 (x-2)/|x-2|

A:  This question is not too bad if you know what the function (x-2)/|x-2| looks like graphically.  But, let’s say you don’t.

We are going to “talk our way” through this problem to help solidify the concept of a limit.

If you plug in 2 to the function, you are finding the value of the function when x=2.  This is important, and related, though it is not the limit.  This is even sometimes a skill used to help us find the limit, but it is still not the limit.  Sometimes the function value is equal to the function limit, which can also be confusing, but not all the time.  Let’s find the value of the function when x=2:

(2-2)/|2-2| = 0/0 = undefined.

Okay.  This function is undefined when x=2.  This means there is a hole, or an asymptote, or a break or a jump or some disruption in the continuity of the function. Continue reading Limit Example

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One-sided Limit Example

Q:  Find the one-sided limit (if it exists):

limx→-1–     (x+1)/(x4-1)

A:  So we need to find the limit of this function (x+1)/(x4-1) as x approaches -1 from the left.  Remember, from the left means as x gets closer and closer to -1, but is still smaller.

The concept: What is happening to this function as x = -2, x = -1.5, x = -1.1, x = -1.0001, etc…

We test first and plug -1 into the function: (-1+1)/((-1)4-1) = 0/0

Whenever you get 0/0, that is your clue that maybe you need to do “more work” before just plugging in or jumping to conclusions.

So, let’s try “more work” — usually that means simplifying.  I see that the denominator can factor.  We have:

(x+1) / (x4-1) = (x+1) / [(x2-1)(x2+1)]

Let’s keep factoring the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

Now, it appears there is a “removable hole” in the function.  This means, we can remove this hole by reducing the matching term in the numerator with the matching term in the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

= 1 / [(x-1)(x2+1)]

Notice that hole exists when x = -1 (and it was removable! This is good news for us since we are concerned with the nature of the function as x approaches -1)

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Integration: Sines, Cosines and U-Substitution

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:

∫cos5(x)sin4(x)dx

∫cos(x)*cos4(x)*sin4(x)dx

∫cos(x)*(cos2(x))2*sin4(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx

Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

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Integration: Disk (Washer Method) vs Shell Method

Here is an example to help you understand and visualize the difference between the disk method and the shell method.  I will do the same problem twice: once using disk method and once using shell method.  Keep in mind: not all problems are equally as easily solved with both methods — that’s why we have multiple methods!  Some problems may be easy to do with the shell method but nearly impossible with the disk method, or vice versa.

If you haven’t read the blog posts that discuss the basic differences between the disk method and the shell method, read those first:

To get the most out of this problem, grab a pen and paper and do the problem along with me: draw each picture and write each equation.

Q:  Find the volume obtained by rotating the area contained by y = √(x), y = 0, and x = 4 around the y-axis.

A:  First, draw a picture of the area that is to be rotated.

Visualize the rotation.  Actually give the shape a name to help solidify it in your mind.  I think of this shape as a cinder cone:

Obviously this isn’t a perfect visual match, but giving a real image to an abstract shape really helps with mentally processing the steps.

Method 1:  Disk (washer) Method.  Remember, the disk and washer method are the same thing.  In the disk method, we visualize stacked circles (or pancakes, as I like to say).  The washer method is the same stacked pancakes with holes (like washers or donuts).

I like to complete every problem with the same thought process.  Follow these steps for all volume/integration problems and you will get the hang of it!

Step 1.  Determine if this is a dx or a dy problem.

Which way are we stacking our pancakes?

Pancakes are being stacked vertically, so this is a “dy problem”.  This means the limits of integration and the equations used will all be in terms of y.

?? ____?____  dy

Step 2.  Find the limits of integration.

Where does the pancake stacking start?  Where does it end?  Remember — in terms of y since this is a dy problem.

Pancakes start at y=0 and end at the place where y = √(x) and x = 4 intersect (the top of the cinder cone).  These bounds intersect at the point (4, 2).

So, pancake stacking starts at y=0 and end at y=2.

02 ____?____  dy

Step 3.  Find the equation of the areas that are being stacked.

We’ve pre-decided that we are going to use the disk (washer) method.  There are definitely washers happening, because there is a big outer circle minus a hole (to create the cinder cone).  So, we need to stack outer pancakes minus inner pancakes (big circles minus small circles):

Area = πR2 – πr2

Let’s start with the area of the big circles, which I’ve called πR2.  What is the radius, R, of the big circles?  Is it changing throughout the problem or is it constant?

The radius of the larger circles is constant.  The larger circles have a radius of 4 throughout the entire cinder cone:  R = 4

Now look at the smaller circles.  Is the radius, r, constant or changing?

Notice the radius of the smaller circles is changing.  This radius is a horizontal distance, starting from the y-axis and moving on out.  The radius is x: r = x.

BUT WAIT…. remember, this is a dy problem.  All equations need to be in terms of y!  So, using r = x will not help.  We need to find a way to represent x in terms of y.  Fortunately, we have the equation to help: y = √(x), so, x = y2

Good.  So, R = 4 and r = y2

Our equation is now complete and ready to solve:

02 π(4)2 – π(y2)2 dy

I will leave the solving to you… but as a final answer I get 128π/5 (approx = 80.425)

Now… Are we ready to solve this same problem using the shell method?

Method 2:  Shell method.  Instead of visualizing stacked pancakes to create our cinder cone, we will visualize stacked “Russian Dolls” (cylinders).  We are going to stack these cylinders so tightly together that the lateral area of the cylinders will stack to create volume.

Back to our steps:

Step 1.  Determine if this is a dx or a dy problem.

The dolls are being stacked inside of each other and on outward, expanding along the x-axis.  This is a “dx problem”.  This means the limits of integration and the equations used will all be in terms of x.

?? ____?____  dx

Step 2.  Find the limits of integration.

Where does the cylinder stacking start and where does it end?  In other words: what is the radius of our smallest cylinder and what is the radius of our largest cylinder?

The smallest cylinder has a radius of x=0 and the largest cylinder has a radius of x=4

04 ____?____  dx

Step 3.  Find the equation of the areas that are being stacked.

We are stacking lateral areas of a cylinder, which has equation: 2πrh (r is the radius of the cylinder, h is the height).

Let’s figure out the radius, r, of a random cylinder/doll in our shape.  The radius does change, so it is a variable.

But, it does not necessarily depend on the functions.  The radius is simply an x-value that continues to grow until we hit the wall of x = 4.

So, r = x.

Now, let’s figure out the height, h, of a random cylinder/doll in our shape.  The height is changing and it is definitely affected by the functions.  The height is the y-value of the bounding function: h = y…. BUT WAIT… remember, this is a dx problem — no y’s allowed. So, use the equation: y = √(x).  Therefore, h = √(x)

So, we have:

04  2π(x)(√(x)) dx

I will leave the solving to you… but as a final answer I get 128π/5 (approx = 80.425) — no coincidence that this is the same answer obtained by method 1!

Same shape, two methods, same answer.  Phew.

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Quadratics: Standard Form to Vertex Form

Q:  Write the equation of y = -x2 + 2x + 2 in the form of y = a(x-h)2 + k

A:  To do this, we are going to use a strategy called “completing the square” — a fairly complicated algebra 2 concept.

We are currently in the form y = ax2+bx+c and we want to get to y = a(x-h)2 + k

First, let’s establish a, b, and c in our equation: y = -x2 + 2x + 2

a = -1, b = 2, c = 2

I will walk you through the steps on how to do the problem.  At the end, I will provide some explanation behind the concept and the “why”.

Step 1)  Divide the entire equation by “a”

So, divide everything by -1:

y = -x2 + 2x + 2

y/ -1 = (-x2 + 2x + 2) / -1

And simplify to get:

-y = x2 – 2x – 2

Step 2)  “Complete the Square”

This step involves finding what number to add (or subtract) into the equation that will make the x terms factor nicely.  To find this number, we follow a simple pattern:

Take 1/2 of the coefficient in front of the x term and then square it.

The coefficient in front of x is -2.

1/2 * -2 = -1

Square -1 to get +1.

The number we need to “Complete the Square” is +1.

Add this number to both sides of the equation (remember to do it to BOTH SIDES to keep the equation balanced).  Also, add in by the x’s just for ease, like so:

-y + 1 = x2 – 2x + 1 – 2

So, our modified equation is:

-y + 1 = x2 – 2x + 1 – 2

Step 3)  Factor the x terms:

We are now going to factor the part of the equation I’ve highlighted.  The part we factor involves the x’s and the “complete the square” number that we added to the equation:

-y + 1 = x2 – 2x + 1 – 2

Factoring x2 – 2x + 1 gives (x – 1)(x – 1) OR (x – 1)2

So, we now have:

-y + 1 = (x – 1)2 – 2

Step 4) Isolate y

We have finished the hardest part (completing the square)!  Now, we just solve for y and we are done:

-y + 1 = (x – 1)2 – 2

Subtract 1 from both sides:

-y = (x – 1)2 – 3

Multiply everything by -1:

y = -(x – 1)2 + 3

And there it is!  We have taken the original equation: y = -x2 + 2x + 2 and re-written it in a different form: y = -(x – 1)2 + 3

Some logic behind the process:

Completing the square is the process of figuring out “what number” is needed to add (or subtract) in the equation so that it will factor easily into something like (x-h)2.  Once we determine the number needed, we add it to both sides of the equation to maintain balance.  Remember, we aren’t altering the actual equation, we are just changing its appearance.  Once we found the correct number, the equation will factor the way we need it to.

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Multiplying Polynomials

Q: Multiply -9m(2m2 + 6m – 1)

A:  When adding and subtracting polynomials, you can only combined like terms.  This is not the case with multiplication.  You can multiply unlike terms together.

-9m(2m2 + 6m – 1)

In this problem the whole quantity in the parentheses is being multiplied by -9m.  So, -9m needs to multiply each term:

-9m(2m2 + 6m – 1) = -9m*2m2 + -9m*6m + -9m*-1
Now we need to clean up the expression on the right:

-9m*2m2 + -9m*6m + -9m*-1 ….

-9m*2m2 can be thought of as -9*m*2*m*m = -18m3

(multiply the coefficients together, and multiply the m’s together)

-9m*6m = -54m2

-9m*-1 = 9m

So, final answer cleans up like:

-9m(2m2 + 6m – 1) = -18m3 – 54m2 + 9m

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Subtracting Polynomials

1:  Subtract: (-4y2-3y+8) – (2y2-6y+2)

A:  Here we are just asked to clean things up and combine like terms: basically, add apples with apples and oranges with oranges, but don’t accidentally add apples with oranges!

y2 terms get combined with y2 terms
y terms get combined with y terms
constants (lone numbers) get combined with constants

So, let’s use color to highlight the like terms:

(-4y2-3y+8) – (2y2-6y+2)

How many total y2 terms do you have? -4y2 – 2y2 = -6y2

How many total y terms do you have? -3y – -6y = -3y+6y = 3y

How many total constant terms do you have? 8+2 = 8 – 2 = 6

Did we forget anyone?  Did anyone not have a match that we need to bring along?  Nope! Everyone from both sets of parentheses is accounted for!

(-4y2-3y+8) – (2y2-6y+2) = -6y2 + 3y + 6

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A:  Here we are just asked to clean things up and combine like terms: basically, add apples with apples and oranges with oranges, but don’t accidentally add apples with oranges!

x3 terms get added with x3 terms
x2 terms get added with x2 terms
x terms get added with other x terms

So, let’s use color to highlight the like terms:

(3x3+2x2-5x) + (-4x3x2-8x)

How many total x3 terms do you have? 3x3+ -4x3 = -1x3

How many total x2 terms do you have? 2x2+x2 = 1x2

How many total x terms do you have? -5x+ -8x = -13x

Did we forget anyone?  Did anyone not have a match that we need to bring along?  Nope! Everyone from both sets of parentheses is accounted for!

(3x3+2x2-5x) + (-4x3x2-8x) = -1x3 + 1x2 13x

Since we tend not to write the “1”, you might see the answer displayed as:

x3 + x2 13x

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Betting and Errors in Poker

If we break down betting to the very basics, there are two types of bet, or two reasons you bet:

• Value Bet: A bet with the goal of getting worse hand to call
• Bluff: A bet with the goal of getting a better hand to fold

To determine the appropriate type of bet to make, you need to ask yourself “What type of errors are my opponents (or is this particular opponent) making?”

If poker were a game like chess, we would analyze which types of bets are best based on your hand, your position, your chip stack, the board, etc…  In chess, moves can be analyzed regardless of your opponent.  In poker, this is not the case.  Whether you should be value betting or bluffing hinges on the errors being made by your current opponent(s).

Are you opponents putting chips in the pot with worse hands?  Are your opponents folding too often?  These questions help determine your reasons for betting.

Types of Errors

First, let’s look at the definitions of Type 1 Errors and Type 2 Errors and what these errors look like in the world of statistics.

Example:  You want to know if you are pregnant and you take a pregnancy test.  There are 4 potential outcomes:

• You are not pregnant and the test comes back negative.
• You are not pregnant and the test comes back positive.
• You are pregnant and the test comes back negative.
• You are pregnant and the test comes back positive.

Two of the above outcomes lead to correct results.  Let’s view this in a table:

 Not Pregnant Pregnant Negative Result Correct Result Positive Test Correct Result

The other two outcomes, which are blank in the table above, lead to errors (incorrect results). These errors are defined as:

• Type 1 Error: You are not pregnant and the test comes back positive – a false positive.  A Type 1 Error is the assertion of something that is absent.
• Type 2 Error: You are pregnant and the test comes back negative – a false negative.  A Type 2 Error is the failure to assert something that is present.

In table format:

 Not Pregnant Pregnant Negative Result Correct Result Type 2 Error – False Negative Positive Test Type 1 Error – False Positive Correct Result

In the pregnancy example, think about the consequences of each error type as it pertains to an individual (or a society). What are the implications of Type 1 Errors? What are the implications of Type 2 Errors?

Betting Errors

Looking at the basics of betting, we can create a similar table to model a simplified situation.  In this situation, you bet.  Your opponent either has a better hand or a worse hand, and either calls or folds.  See the errors that can be made by your opponent below:

 Opponent has a better hand Opponent has a worse hand Opponent Calls No Error Type 2 Error Opponent Folds Type 1 Error No Error

Bluff

When you bluff, you are counting on your opponent to make a Type 1 Error.  You are asserting something that is absent (a strong hand) and your opponent falsely believes you and folds.

If your opponents are waiting for strong hands and folding too often, they are making Type 1 Errors.

Value Bet

When you value bet, you are counting on your opponent to make a Type 2 Error.  You fail to assert strength (making your opponent sense weakness) and your opponent falsely believes you and calls.

If your opponents are calling too much or seeing too many rivers, they are making Type 2 Errors.

Are you helping your opponents correct their errors?

You need to take advantage of the types of errors your opponents are making.  As well, your bets should not encourage your opponents to correct their errors.

For example: drastic over-bets encourage players who make Type 2 Errors to play correctly and fold when otherwise they would have called a standard-sized bet with a weaker hand, giving you lots of value.

And, a player who makes Type 1 Errors will fold to standard-sized or over-sized river bets, but be encouraged to play correctly by calling smaller-sized bets.

Figure out which type of errors your opponents are making and then bet accordingly.