Let’s take a look at the term function.
In everyday language, the word function could be used like this:
you could say the income that a person earns as an adult
is a function of the level of education that person attains.
So what does that mean? That means that a person attains a level of education
and then from that results an income,
and that is fine in everyday language,
but when you talk about a function in mathematics,
the definition is much more specific.
There is one particular criteria that a function must have.
Here is what it is: for each item in the first group, there must be one
and there must be only one item that results from that in the second group.
Alright, let’s take a look at some illustrations from everyday life.
This first one is an illustration of a situation that is not a function.
We will take a group of people. Let’s call them Larry, Michael, and Joel.
These will be the items in the first group.
The function that is going to produce the items in the second group
would be to list their phone numbers.
Okay, so let’s see. Take Larry and we will list his phone number: 642-8729.
Then we will take Michael, and when we ask him his phone number
he says: 931-1846 or 766-6956.
When we ask Joel he says his phone number is: 862-4193.
Why is this not a function? It is not a function because this one element in the first set
maps to two elements in the second set,
and that violates the definition of a function.
So this situation does not represent a function.
Alright, let’s look at another illustration.
We use the same three people,
but we will use a different function.
So this time the function will be list the social security numbers.
Okay so Larry’s social security number is 438-61-2915.
Michael’s social security number is 434-64-3928.
Joel’s social security number is 435-19-2003.
Does this represent a function?
Yes it does because for each element in this set there is one
and only one element in this set.
Let’s take a look at one more example.
Another everyday situation that you are familiar with
is the fact that a person has a birth date.
In this case, we will use month and date. So this is person was born on.
Okay, so Larry was born on October the 19th.
Michael was born on January the 24th.
It turns out that Joel was also born on January the 24th.
The question is: does this represent a function? Well let’s see.
For each element in the first set, there is one
and only one element in the second set.
It is perfectly okay that these two elements here to
match to the same element in the second set.
Yes, this represents a function.
The elements in this set are called the domain.
The elements in this set are called the range.
Let’s take a look at some examples.
Are these functions? If yes, state the domain.
The domain is the set of all elements in the first set.
Let’s look at part a. We have three ordered pairs: (2,4), (2,6), and (2,8).
Let’s draw this. The 2 in the first ordered pair maps to 4.
The 2 again in the second ordered pair maps to 6.
The 2 in the third ordered pair maps to 8.
What we have here is one element but it maps to three different elements here.
This violates the criteria for a function that says that each element here
maps to one and only one element here.
So no, this does not represent a function.
In part b, we again have three ordered pairs: (2,1), (3,1), and (5,7).
The 2 maps to 1, the 3 also maps to 1, and then the 5 maps to 7.
This is equivalent to two different people having the same birthday,
and that is perfectly fine. That meets the definition of a function.
Yes, this is a function. Now because the answer is yes,
we have to state the domain.
We are going to state the domain here as a set,
and it is going to contain three elements: 2, 3, and 5.
In part c we have a slightly different notation. We have a set defined here.
This is read: the set of all ordered pairs, (x,y),
such that y is equal to 2x minus 1.
So the rule here, or correspondence,
is take an element in the domain, multiply it by 2,
subtract 1, and what you get is the resulting element
in the range, which is y. Alright, so let’s think about this.
Well in this case, x can be any real number.
There is no reason why we could not use any number: positive, negative, or zero.
We put it into here, we multiply by 2, subtract 1, and what happens?
The result is we get a real number.
What will not happen is that when we put an x in here,
we will never get 2 different y’s for that x. So, this is a function. Yes.
The domain, or the choices of the elements in the domain, are going to be all real numbers.
If we write this in interval notation, we can write it as negative infinity to infinity.
In some cases, we may just use this capital R symbol. Okay, now let’s look at part d.
What happens here from part c to part d, is that we realize that overtime,
this notation for part c is very cumbersome.
So what happens is we tend to drop part of it,
and we just use this description of how you get from x to y. Alright so let’s look.
Again, there is no reason why any value of x would not work here.
You put it into here, square it, and multiply by 2.
Then you multiply it by 7, add it this, and add 9. Follow the rule. What happens? You get a y.
What happens here is you put in an x, and you get one and only one y.
So yes, this is a function. The domain again is going to be
either one of these two notations for all real numbers.
Alright, let’s take a look at part e. You should be suspicious right away when you
see a variable in the denominator.
There is going to be some sort of restriction on the domain. What is it?
Well x minus 6, what? x minus 6 cannot equal 0.
So that means that x cannot be 6. We see there is a restriction on the domain.
But what about this is a function?
Well let’s see. As long as we do not use 6, we are okay,
because any real value that we put into here will give us one value of y.
Again, yes this is a function, but the domain is not going to be all real numbers.
It is going to be either this notation, or depending on the situation,
you may be able to just put x is a real number not equal to 6.
It depends on what the directions are and
what you are trying to do with this. Alright, let’s look at a few more.
y squared equals x minus 4. Is this a function?
Well let’s see. Let’s pick a value for x,
and let's see what happens, and what y we get.
I am going to pick something for x that gives
this right hand side a result of a perfect square. So I am going to pick x is 13.
So what happens is, then I get y squared equals 13 minus 4, which is 9,
and which means I get y is plus or minus 3.
Now if you go back and look at what we did with mapping,
that means that this x value of 13 maps to 3, but it also maps to negative 3.
What do you know about that?
You know that what you have here is an element in the domain
that maps to two different elements in the range,
and this violates the definition of a function.
Therefore this does not represent a function.
Now let’s just stop a minute and think about why this is the case.
This is the case because when we put a value here, in here for x,
we get one value here. But in order to solve for y, we get two values.
So what does that tell you? That tells you that anytime you have a situation
where x in the independent variable, y is the dependent variable,
you are going to get a situation when you have y squared,
where this is not a function. This is because every time you do this and
you take the square root of both sides, you are going to get two values for y
that go with the same x. You can start looking for this
and anytime you see this situation, you will know it is not a function.
Okay that brings us right into the next one.
We have x squared plus y squared is 9.
So what do you know? What did you learn from f?
You learned that in this case when you substitute a value in for x,
the resulting values for y, each time you put in one x, you will get two values of y.
What happens is you know right away that no, this is not a function.
Okay, in part h we have a restriction on the domain.
2x minus 1. What is the restriction? Think about it.
2x minus 1 cannot be negative, which means it has to be what?
It has to be greater than or equal to 0.
Which means that 2x has to be greater than or equal to 1.
Which means x has to be greater than or equal to 1/2.
So let’s see. Let’s pick a value in this domain.
Okay so let’s pick something like, oh I don’t know, 6.
Okay so when x is 6, 6 times 2 is 12, minus 1 is 11.
We get the square root of 11. We get one value.
So for this one x we get one y, and this is fine.
Yes a function. And what is the domain?
You could write the domain like this,
or you could write it in interval notation.
You want to use a closed interval here and infinity here.
This is the domain of this function. Alright, one last one.
This expression here has not only a variable under the radical,
but the radical is in the denominator.
So let’s look at the domain to start off with.
Well first of all, x plus 3, it cannot be 0, and it also cannot be negative.
So what does that leave? Well that means that x plus 3 must be positive,
so x must be greater than negative 3.
Okay so let’s pick something greater than negative 3 like say 10.
10 plus 3 is 13, so you get 4 divided by the square root of 13,
which is one value of y. So for each x,
you are going to get one and only one y.
So yes this is a function. Here is the domain:
x is greater than negative 3.
Notice here that this is an open interval,
not including the negative 3, and so here is the domain.
Alright, let’s look at a few more things about functions.
Let’s talk for a second about function notation.
Now if x represents an element in the domain
and y represents an element in the range, then what you have is
that the y you get is simply a function of the x that you choose.
Okay, now let’s say that again. The y that you get is a function of the x that you choose.
So y is a function of x. Okay so say that to yourself a few times.
y is a function of x. Then think about it over and over and over again.
Just like everything else, we come up with a shortcut.
What does it look like? It looks like this.
y is a function of the x that you choose.
Okay, and so this becomes the standard notation used to represent a function.
So that means that in an equation, in place of y,
when you know that this equation represents a function,
you can simply replace this with f of x.
It is very useful notation. Let me show you what I mean.
In example two, the directions are: for the function defined by, and let’s put this
in a box so it is easy to spot.
Here is the function we are going to work with.
This is a function, and we know it is a function
because right here instead of y, it says f of x.
Okay so here is the function, we want to find the following:
f of negative 1. Well this means that x is going to be
replaced with negative 1.
So everywhere in this expression, I replace x with negative 1.
So we have 2 times negative 1 squared, minus negative 1, plus 3.
Then I am going to evaluate the expression.
Be careful here with signs. Negative 1 squared is 1. So this would be 2,
then this would be plus 1, and plus 3, which is 6.
So what we have is that f of negative 1 is 6. So what does that mean?
That means when x is negative 1, the resulting y is 6.
Alright, now this time it says replace all the x’s with the opposite of x.
So we would have 2 times the opposite of x squared,
minus and in place of x I am going to put the opposite of x,
and then plus 3. Okay, the opposite of x squared is x squared.
So I get for the first term 2x squared.
Then here I get plus x, and then plus 3.
So what that means is if I put in the opposite of a value of x,
this is the resulting expression that I would get.
Alright now, part c is different. Part c says I am going to
take the opposite of the whole thing.
So that means what you are going to do is
you are going to take the entire right hand side here,
and replace it here. We would get
the opposite of then the entire expression.
We will clean it up a little bit, distribute this negative.
We will get negative 2x squared, plus x, and then minus 3.
Okay, part d says now, in place of the original x’s,
I want you to put the binomial x plus h.
Alright, so I still have 2, then I have x plus x squared,
minus, and then I need parenthesis, x plus h, and then plus 3.
Okay, so I have replaced the original x
with x plus h here and x plus h here.
Now I could simplify this, but what I am going to do
is I will just hold that until the next example.
In part e, we want to find f of x plus h, as we did in part d,
then we want to subtract the entire function, f of x,
and then divide by h. Alright, let’s do this one step at a time.
First of all, let’s find f of x plus h again.
This means put x plus h in for this x and this x.
So I have 2 times, x plus h quantity squared,
minus x plus h, plus 3. Then that takes care of this part right here.
Now I need to subtract the function.
So I am going to subtract 2x squared minus x plus 3.
Then I am going to take all of that and divide it by h.
Okay, so that equals, I am going to square this term.
Now x plus h squared, let’s go up here and do this.
This would be x plus h squared is x squared plus 2xh plus h squared.
That is this term. I want to multiply all three of those terms by 2.
So I would get 2x squared plus 4xh plus 2h squared.
That takes care of this part. The most common error on this problem
is right here, is to lose this 2 here
or to lose this 4 here. So watch out when you do this. Very common error.
Alright now I am going to distribute
the negative here, copy over the plus 3,
and now distribute the negative all the way across.
Take your time when you do this problem.
It is really all about signs and handling the algebraic expression,
and then of course all of this is divided by h.
Okay, so what happens? Well let’s see.
2x squared minus 2x squared is 0.
Then minus x and plus x is 0, and 3 and minus 3 is 0. If you did it right,
this is what is going to happen every time.
All the terms that have no h’s are going to end up canceling,
and what we have left is 4xh plus 2h squared minus h, all divided by h.
So what I have is an h as a factor in all three terms in the numerator.
I am going to factor out the h in the numerator.
That will leave me 4x, and plus 2h minus 1.
Again, very common error here is when you factor h out of this term,
you do not get 0, you get negative 1.
Then the whole thing is over h, and then now you have h divided by h.
The final simplification is to get 4x plus 2h minus 1.
This particular expression turns out to be
something very important in calculus.
This is a skill you want to practice and get really good at.
Let’s look at a few more things.
The domain of a function is the set of all
values of x for which the function is defined.
It is often very helpful to classify a function to determine its domain.
The first class of functions we will look at is polynomial functions.
So here is the definition: the function f of x, a sub n x to the n,
plus a sub n minus 1 x to the n minus 1,
plus a sub n minus 2 x to the n minus 2,
plus and it continues in the same manner then, plus a sub 1 x,
which of course is to the 1, plus a sub 0, which is times x to the 0,
is a polynomial function of degree n where n is a nonnegative integer,
and the numbers a sub 0, a sub 1, a sub 2, etc., to a sub n are real numbers.
The domain of every polynomial function is negative infinity to infinity.
Now do not be intimidated by all this notation.
What this simply means is that a sub n is the coefficient that goes with
x raised to the n power. a sub n minus 1 is the coefficient that goes with
x raised to the n minus 1 power, and so forth.
A second class of functions is rational functions.
A rational function is a function of the form:
f of x equals g of x divided by h of x
where h and g are polynomial functions
such that h of x does not equal 0.
So the domain of a rational function is
the set of all real numbers x such that
h of x does not equal 0.
The third class is root functions.
The function f of x equals the nth root of g of x
is a root function where n is an integer,
such that n is greater than or equal to 2.
Now there are two possibilities for the domain.
One: if n is even, the domain is then the solution to the inequality
g of x is greater than or equal to 0.
The second case is where n is odd, then the domain is
the set of all real numbers for which g of x is defined.
Now let’s look at some examples.
Example three: Classify the given functions as a polynomial function,
rational function, or root function, and then find the domain.
Write the domain in interval notation.
Part a is the function f of x equals 2x minus 11.
This is a polynomial function. It has degree one.
The exponent of x here is 1. So we classify it as a polynomial function.
Because it is a polynomial function,
the domain, in interval notation, is negative infinity to infinity.
Alright, let’s look at part b. f of x equals 3x minus 1 divided by 3x plus 1.
We can see that this function is of the form g of x over h of x,
and so this is a rational function.
If there is a restriction on the domain,
we find it by setting h of x, the denominator, equal to zero.
What I am going to put here is that h of x does not equal to 0,
and then I will solve. I subtract 1 from both sides, and divide by 3.
I see that x is not equal to 1/3. But of course I have to be careful
because when I subtracted 1 from both sides I should get negative 1 here,
and negative 1/3. Always check before you go on.
Now in interval notation, I am going to write this as
negative infinity to negative 1/3 union negative 1/3 to infinity.
Let’s look at a few more. Part c: f of x equals the fourth root of 2 minus x.
We can see that this is a root function where n is 4.
So let’s classify this as a root function.
Because 4 is even, the domain will be the values of x
where 2 minus x is greater than or equal to 0.
Okay, so we add x to both sides and we get x is less than or equal to 2,
and in interval notation we get this for the domain.
Part d: f of x equals the seventh root of 5x plus 9 is also a root function.
It has the same form but the difference is that here, this is the seventh root.
So 7 is odd and in that case, without doing any work,
we know that the domain is negative infinity to infinity.
Let’s look at one more. Part e: f of x equals the fifth root of x plus 2 divided by x minus 9.
When you look at this you see the fifth root,
and so you classify this as a root function.
Because n is odd, you might think originally
that the domain would be all real numbers.
But be careful, look at this. Okay, It is not possible here for x minus 9 to equal 0.
So we have to exclude x equals 9 from the domain.
So the domain of this function is this.
Okay, now we move on to a second set of topics relating to functions.
Let’s take a look at graphs of functions.
Remember that the definition of function says that
one element in the domain is mapped to one and only one element in the range.
Now how does that apply to the graph of a function?
Well, let's see. We have eight graphs here,
and we want to determine if these are graphs of functions.
So let’s see what we can come up with that
might help us to do this efficiently.
Well, let’s see for example any value of x that I select here will map to one value of y.
Okay, so for example if I pick an x here,
then there is the ordered pair (x,y) on the graph.
You can see that on this particular example,
any x you pick will have one value of y.
So yes, this is a function. Okay, let’s look at the second one.
What happens here is that you have only one value of x.
This is a vertical line and the value of x is the same on this line,
and so what happens is you might have a point
here that is going to be (x,0), and then up here somewhere you have exactly
the same x but maybe the y value is 5.
So what you can see here is that a vertical line
does not represent the graph of a function.
So the answer to b is, is this a function? No.
But let’s get a little bit more out of this.
Let’s realize here that a vertical line is never going to represent a function.
Not only that, we could use a vertical line to help us
determine whether or not a graph is a function.
So let’s go back to a and see how this would work.
If you visualize vertical lines anywhere on the graph
you can see that any vertical line that you would draw
would touch this graph only one time.
What happens is, here, when you draw a vertical line,
it touches this graph in an infinite number of points, all the points on the line.
What results here is what is called the vertical line test.
Okay, we can just call this VLT for vertical line test. Here is how it works.
If you can draw a vertical line that crosses a graph more than once,
or touches more than once, then that is not the graph of a function.
Okay, so let’s try it on part c.
Notice that no matter where we draw a vertical line,
it touches the graph only one time.
So yes, this horizontal line represents the graph of a function.
Okay, part d. As we draw vertical lines,
we can see that this vertical line crosses,
or any vertical line crosses the graph only once.
Yes, this represents a function. Okay, what about a circle?
Well anywhere between here and here that we draw a vertical line,
you can see the vertical line is going to cross the graph twice,
and that is because for the same value of x you have two different y-values.
That violates the definition of a function.
So no, circles are not going to represent graphs of functions.
Once again, vertical lines. Anywhere that you draw the vertical line,
you get only one cross. This is the graph of a function.
Oops, what happens here? Anywhere from here on out,
when you draw a vertical line, the vertical line crosses the graph twice.
So you have, right here, you have x’s with two different y-values.
This does not represent a function.
The last one: any vertical line crosses only once,
and so yes, this is the graph of a function.
Alright, let’s take a look at more graphs of functions.
Let’s use the graph of this function, f, to answer the following questions.
Be very careful here about the notation that you use.
Part a, what is f of 4? Alright, now what does this mean?
This means that we want x to be 4,
and we are looking for the y-value on the graph when x is 4.
In other words, we are looking for the value of the function at 4.
Okay, when x is 4, we look at this area right here.
x is 4 anywhere up and down this line.
On the function we see here that x is 4 when y is 3. So f of 4 is 3.
Do not put the ordered pair (4,3). Just put 3, the y-value.
Okay, part b is f of negative 4 positive or negative?
Well f of negative 4 is a y-value.
It is the y-value of the function when x is negative 4.
Okay, let’s see. x is negative 4 anywhere up and down this line,
and on our specific graph, right there x is negative 4.
You can see that that value is below the x-axis,
and so the y-value here, it looks like it is approximately negative 1.
But we were not asked for the value,
we were just asked if it is positive or negative. So we are going to put negative.
Again, approximately negative 1, but that is the correct answer for the specific question.
Alright, part c: how often does the line x equal 2 intersect the graph?
Well, what does the line x equal 2 look like?
Well, x equal 2 is going to be a vertical line.
Let’s see how many times does it intersect the graph?
Once. Think about the vertical line test and remember
that since this a function that is exactly what you would expect:
a vertical line would only intersect this graph once, and that is what happens.
Okay, the domain. This is the set of all x’s for which the function is defined.
The smallest value of x is negative 5.
As we read from left to right, we see the largest value of x is 9. Both endpoints are included.
In interval notation, we are going to write the domain this way.
Okay the range is the set of all y values for which the function is defined.
So as we read from bottom up to top, smaller to larger values of y,
we can see that that the smallest value of y is negative 2
here and negative 2 here. So that is the smallest value.
The largest value of y is 4. So the range goes from a value of negative 2 to 4.
Again both endpoints are included and so I put brackets.
Okay, part e: for what values of x does f of x equal 4?
Okay now, this number right here, this is a y-value.
So what this says is when y is 4, what is x?
Okay, well y is 4 anywhere along this line that goes right through here.
You can see that the only x-value on the function where y is equal to 4 is 2.
So we are going to write down the number 2.
It would not be correct to write the ordered pair (2,4).
The question says what is x when y is 4. Answer: 2.
Okay, part f is kind of a similar question.
It says when f of x is 3, what is x? Okay, that means when y is 3.
Okay, y is 3 right along this line,
and you can see that on our function
there are two places where the graph has a y value of 3.
Here when x is 0, and also when x is 4.
Okay, so I am going to list these two answers.
Of course you have to read the directions on a problem,
and be careful that you answer the question exactly as it is asked.
But in this case I am just going to
list the two values 0 and 4, no parenthesis.
This is not an ordered pair, these are values of x, this 0 and this 4.
Okay, last question: what are the x-intercepts?
Now x-intercepts are the actual values of x
where the graph crosses the x-axis.
Sometimes intercepts are listed as ordered pairs,
but in this when you are just asked to list the x-intercepts,you list just the x-values.
So we would have an intercept here
at negative 3, one at 6, and another one at 9.
So I would put negative 3, 6, and 9.
Okay, if I were to list these as ordered pairs,
I would put (-3,0), (6,0), and then (9,0).
The y-intercept. Of course this is a function
so there is only going to be one y-intercept.
Here it is: it occurs at the point (0,3),
and so I am going to list the y-intercept, and the y-value is 3.
Alright, let’s take a look at a graph of another function.
Let’s answer these questions about the function
f of x equals 3 x squared minus x minus 2.
This is the function we are going to work with.
Part a asks is the point (2,8) on the graph of f?
This means is f of 2 equal to 8? Well, let’s test.
f of 2 would equal 3 and then 2 squared minus 2 minus 2.
Let’s check that. I substituted 2 into here and here.
So this is f of 2 and my question is: is it equal to 8?
2 squared is 4. 4 times 3 is 12.
So I get 12 minus 2 minus 2, which is minus 4,
and I get 8. So what that means is that f of 2 is equal to 8,
which means that the point (2,8) would be on the graph of this function.
So the answer is yes.
Part b: if f of x is negative 2, what is x?
Okay, this means that y is equal to negative 2.
Now y is f of x, y on the graph is f of x here.
So I am going to substitute in place of f of x negative 2.
So I am going to have negative 2 equals, and then this expression,
3 x squared minus x minus 2.
Okay, wen I add 2 to both sides, I get 0 here and 0 here.
I am going to reorder this, make it easier to look at.
I am going to write 3 x squared minus x equals 0.
I want to solve for x. I am going to factor an x out. That leaves me
3x minus 1 equals 0 for the other factor.
I am going to set each factor equal to 0.
Don’t lose this factor, that is a common error.
I am going to get either x equal 0 or 3x minus 1 equal 0,
which means x equals 1/3. So what that means is that when f of x is negative 2,
there are two values of x: 0 and 1/3.
This is not an ordered pair, this is a listing of two x-values.
Then the c part is to state the domain.
Well this function has no restriction on the domain.
There is no reason why you could not use any real number
for x substituted into here and get a value for f of x.
So the domain is all real numbers.
In interval notation, it would be written this way.
Or depending on the directions you could put a capital R symbol for real numbers.
Alright, you need to study these examples carefully,
and be sure that you have mastered these skills.
These are very important skills about graphs of functions.
Part d: list the x-intercepts. The x-intercepts are going to be
the x-values where y, or f of x, equals zero.
So let’s let f of x equal 0, and we are going to write
3x squared minus x minus 2, and what we want to know is
what values of x makes this equal to 0.
Now, this is a quadratic equation
so the first thing we are going to do is attempt to factor it.
Let’s see if it factors easily. It might.
We can see that we need 3x then x,
and then the factors of 2 will be 2 and 1,
and so if we put a 2 here and a 1 here,
and then we want the 3x to be negative
so I put a negative here and positive here.
Let’s check it. That is correct.
We set each factor equal to 0.
So we get 3x plus 2 equals 0. x is negative 2 divided by 3.
Then here we get x minus 1 is 0.
x equals 1. So we have two values of x were f of x equals 0,
and these are our two x-intercepts.
I am going to list them. These are not ordered pairs.
If they were ordered pairs, they would be listed as
(-2/3,0), one pair, and (1,0) as the other pair.
Okay, now the y-intercept is going to be the value of y where x is zero.
So that means I am going to find f of 0.
This is going to be pretty simple because f of 0 is going to be
0 for this term, 0 for this term, and what I have left is negative 2.
That is of course the one and only y-intercept, negative 2.
Alright, part e: if x is negative 1, what is f of x?
Well, this is just another way of asking what is f when x is negative 1.
Okay, and so I am going to substitute negative 1 into here.
Negative 1 in for the other x, and then
I am going to simplify the expression on the right.
This is positive 1 times 3, so this would be 3 plus one minus 2,
which is 4. So f of x is equal to 4 when x is negative 1.
This is, let’s see, I think maybe that should be 2.
It would be 4, I better finish this, 4 minus 2 and so this should be 2.
Always be careful and check your work.
Alright, you need to carefully practice.
Study these examples until you have attained
a very high skill level with this information.
Example four: use the graph of the function to answer these questions.
So here we have the graph of a function,
and part a says is f, the function, increasing on the interval from 0 to 5?
Alright, now let’s take a look at this question carefully.
First of all, this is not an ordered pair (0,5),
this is an interval as x goes from 0 to 5.
So let’s look at where this is on the graph.
Okay, well x equals 0 starts here and x equals 5,
three, four, five, goes to about here.
So what we are doing here is we are looking at the interval from 0 to 5.
When you talk about increasing or decreasing you always,
always, read from left to right.
Do not pay any attention to what the arrows
on the ends of the graph may cause your eye to do. You read from left to right.
Alright, so over the interval from 0 to 5
let’s look at what the graph is doing.
It is decreasing, increasing, decreasing, and then increasing to about right here.
Is f increasing on the interval from 0 to 5?
Clearly the answer is no. It is doing some increasing and some decreasing.
Now, the b part gets more specific.
It says list the intervals, intervals: values of x,
on which f, the function, is increasing.
Then the c part is going to be the same thing but decreasing.
So what we are going to do is we are going to read from left to right,
and we are going to identify all of the intervals
where the function is increasing and decreasing.
Then we are going to back and write the answers down.
Alright, so as I read from left to right,
I can see that the graph is decreasing as x goes from
actually negative infinity to the x value negative 5.
So right here, the function is decreasing,
and then the function turns, and begins to increase.
It increases until x gets to negative 2.
Then as x goes from negative 2 to 1, the function is decreasing.
Then as x goes from 1 to 2, the function is increasing.
Then from 2 to 3, the function is decreasing.
Then from 3 to infinity the function is increasing.
Alright, now let’s see if we can get all these intervals written down.
Remember that these are values of x.
You do not write ordered pairs, you write intervals on x.
Alright so from negative infinity to negative 5, the function is decreasing.
We are going to use open intervals for all of these answers.
Then from negative 5 to negative 2, the function is increasing.
Okay, then from negative 2 to 1, the function is decreasing.
Negative 2 to 1. Then from 1 to 2, the function is increasing.
From 2 to 3, the function is decreasing.
From 3 to infinity, the function is increasing.
Now let’s take a look at exactly what we have here.
We should have all of the elements of the domain except
for the actual endpoints where the function changes from
increasing to decreasing. So we have negative infinity to
negative 5, negative 5 to negative 2, negative 2 to 1,
1 to 2, 2 to 3, and 3 to infinity.
So we have covered all of the intervals.
Let’s take a look at a few more questions.
Part d: determine whether this function is even, odd, or neither.
The simplest way to determine this is to
remember what you know about symmetry.
An even function is symmetric with respect to the y-axis.
An odd function is symmetric with respect to the origin.
So let’s take a look. Look at the purple graph of the function.
Do you see and y-axis symmetry or any origin symmetry?
The answer is no, and so we can say that
this function is neither even nor odd.
Part e: list the values of x at which f has a local maximum.
Well let’s see, where would we find a local maximum?
A local maximum would occur where a function changes
from increasing to decreasing.
So right there, that is a local maximum.
Then you see another one occurs here where
the function changes from increasing to decreasing.
The values of x where this occurs would be listed as
negative 2, and 2. This is not a point,
this is not an interval, these are values of x
where we have a local max. What are these local maxima?
This local max is 4, and this local max is 3.
Notice that I do not put ordered pairs here,
I simply put the y-values because the actual max is the y-value,
or the value of the function at these particular values of x.
Okay, same question but about a min.
So list the values of x at which the function has a local min.
Alright, a local min is going to occur where a function changes
from decreasing to increasing. So one here.
Decreasing to increasing, one here.
Decreasing to increasing, so we are going to have three values
of x where we have a local min occur.
Those values would be negative 5, 1, and 3.
Negative 5, 1, and 3, those are the values of x were a local min occurs.
Now the local min itself here is negative 3.
The local min here is 0. The local min here is also 0.
So we are going to list the negative 3, and then the 0.
Even though this occurs twice, we will list it once.
Practice working with these questions and these examples
until you have mastered the skill.