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) 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

A:  OK.  Let’s look at the left end behavior first:

As  x approaches -∞, what is the function (y-value) doing?

Imagine x=-1000000 (some super large and super negative number, like the idea of -∞), we have:

y=5(-1000000)3+7(-1000000)2-2(-1000000)-1

Don’t do the actual math.  Just think:

Is this number large or small?

Is it positive or negative?

I can look at the x3 term and see that it dominates this function. x2 and x are small peanuts compared to x3. So, in reaity, in polynomials, I can focus on the term of the largest degree:

y=5(-1000000)3+7(-1000000)2-2(-1000000)-1

y=5(-1000000)3

This number gives y = negative and super large.

So, I can jump to conclusions here…

As x → -∞, y → -∞

(As x approaches negative infinity, y approaches negative infinity).

Now, let’s look at the right end behavior:

As  x approaches +∞, what is the function (y-value) doing?

Imagine x=+1000000 (some super large and super positive number, like the concept of +∞), we have:

y=5(+1000000)3+7(+1000000)2-2(+1000000)-1

And, by the same reasoning, we can focus on the term of largest degree:

y=5(+1000000)3+7(+1000000)2-2(+1000000)-1

y=5(+1000000)3 = super large and super positive

So, as x → +∞, y → +∞

(As x approaches positive infinity, y approaches positive infinity)

Note: in this example, y behavior mimicked x behavior, this isn’t always the case!

Visualizing limits vs values

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

limit6

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:

limit1

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?

limit2

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?

limit3

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

Are you ready to try one on your own? Click here! (Don’t worry, I’ll walk you through the solutions too)

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.

So, let’s talk about the limit of the function as x approaches 2:

To find the limit as x approaches 2, we need to make sure the left-handed limit and the right-handed limit both exist and are equal.

I’m going to start with the right-handed limit (remember, this means we are getting closer and closer to 2 from the top-side).  We are going to do this by clever analysis:

When x=3, we get (3-2)/|3-2| = 1

When x=2.5, we get (2.5-2)/|2.5-2| = 1

When x=2.01, we get (2.01-2)/|2.01-2| = 1

See the pattern here?

When x>2, the function ALWAYS equals 1 (tricky function huh?).

So this tells us about the right-handed limit at 2:

limx→2+ (x-2)/|x-2| = 1

Now, let’s trying the left-handed limit at 2, again by analysis:

When x=1, we get (1-2)/|1-2| = -1

When x=1.5, we get (1.5-2)/|1.5-2| = -1

When x=1.99, we get (1.99-2)/|1.99-2| = -1

Again, a pattern: When x<2, the function ALWAYS equals -1.

This tells us about the left-handed limit at 2:

limx→2- (x-2)/|x-2| = -1

Since the right-handed limit at 2 ≠ the left-handed limit at 2, “the limit at 2” does not exist.  The right-handed limit must equal the left-handed limit to have a “complete limit” so to speak.  The concept:  If I walk the path from the left, and you walk the path from the right, we better think we are going to the same place.

Here is a picture of the function just for fun.  See what is happening as x approaches 2 from the right?  See what is happening as x approaches 2 from the left?  And, see what happens exactly at 2?

ABSVALUE.png

 

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)

Now that we have removed that hole, let’s once again try to plug in -1 to see what we get.

1 / [(-1-1)((-1)2+1)]

1 / [(-2)(2)]

-1/4

So, after removing the hole at (x+1), we found the function value when x=-1 is -1/4.

Due to the nature of this function, this means:

limx→-1(x+1)/(x4-1) = -1/4

Since the limit exists as both the right-handed and left-handed limit, it follows that limx→-1–     (x+1)/(x4-1) must also be -1/4.  It ended up not being necessary that we only do a “one-handed limit analysis.”

 

Integration Example: Sines an Cosines

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

∫cos(x)*(1-sin2(x))2*sin4(x)dx = ∫ du*(1-u2)2*u4

Or, rewrite it: ∫(1-u2)2*u4du

Simplify and clean up:

∫(1-2u2+u4)*u4du

∫u4-2u6+u8du

Integrate to get:

(1/5)u5-(2/7)u7+(1/9)u9+C

And, substitute back in u = sin(x):

(1/5)sin5(x)-(2/7)sin7(x)+(1/9)sin9(x)+C

This is only 1 way to solve this problem.  I could’ve picked a different u-substitution and I could’ve likely picked different trig identities to help me along the way.

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.