BAM!!! Mr. Tarrou. In this lesson we are going
to look at three examples that deal with or
involve the intermediate value theorem. This
is what it says. If f is continuous on the
closed interval [a,b]... These notes are for
my Calculus class, but I also cover this in
my Precalculus class. If you cover this in
a book that does not use interval notation,
a closed interval [a,b] means that we are
looking at values along the x axis from a
value of a to a value of b and we actually
using those values of a and b. It is closed.
So, like if the value of b is equal to 5.
We are not going up to 4.999999....., it is
a closed interval so it is including the value
of a and b like 5 in this interval. If it
continuous on a closed interval and f(a) is
not equal to f(b), basically the y value we
get from the function when we plug in a is
not equal the y value that we get when we
plug in b into the equation, and k is any
number between f(a) and f(b)... If k is between
f(a) and f(b), then k is representing a y
value on the graph... then there is at least
one number c in this closed interval of [a,b]
such that f(c) is equal to k. I have function
here that is continuous. It happens to be
smooth as well, but it does not need to be,
just continuous. Let's see what color chalk.
Let's get blue. We have got a function which
is continuous on a closed interval from a
to b. We have our end points on our interval
f(a) is not equal to f(b). I am going to come
through here and....besides partially erasing
my diagram...put in a value of k that is in
between f(a) and f(b). Let's have this line
represent, this line representing this value
of k. Just to highlight that y value. Now
with my particular drawing there is actually
1, 2, 3 places where we can find a 'c' in
the interval...really we would call this what...c
sub 1, c sub 2, c sub 3. There is actually
three c values that I can put into my equation
that is creating this graph where the y value
would be equal to k. That is what this saying
right here. f(c), when I take c and plug it
into my equation...it is probably going to
be a function...I want to say a function,
but it does not...it is not stated in the
intermediate value theorem. But, when we take
c and plug it into the equation we get a value
of k. There is this c guarantee such that
we can plug it into our equation that f that
is making this function, and the value from
the function is equal to k and it is guaranteed
to occur as long as that function is continuous
and k is between f(a) and f(b). I have my
first example, but I have taken up my entire
board giving you the notes so let me clear
this off. Before I go on with my example,
I went ahead and drew a graph, and f, that
is continuous on this closed interval [a,b]
and would fail the vertical line test thus
it is not a function. But since it is continuous
and f(a) is not equal to f(b), then somewhere
there is still a value of k, as long as it
is between f(a) and f(b) such that a c is
guaranteed where f(c) is going to be equal
to k. Ok:) So here we have our example. It
says verify that the Intermediate Value Theorem
applies to indicated interval and find the
value of c guaranteed by the theorem. The
value of c that is guaranteed and we are going
to look for hopefully we find if the intermediate
theorem applies, that f(a) is going to be
different f(b) and that this value...k value
of 12 is somewhere f(a) and f(b). It does
not really matter if our k value is between
the values of f(a) and f(b) if this function
is not continuous on this closed interval.
So you want to make sure that the intermediate
value applies and the first thing you have
to do is make sure that it is continuous on
that closed interval. Our function here, and
I am going to go ahead and write the word
domain because we are going to go ahead and
write the answer for domain. The numerator
of this rational function is a polynomial
and we know polynomials to be continuous everywhere
from negative infinity to positive infinity.
The denominator of this rational function
is a polynomial. So again that denominator
is going to be continuous everywhere...basically
with all real numbers...from negative infinity
to positive infinity. What we always have
to check when our variable is in the denominator
of a rational equation. I need to stop the
word function even though this particular
equation will be. We can't divide by zero.
We need to make sure when checking for domain
that x-1 cannot equal zero, thus x cannot
equal positive one. In interval notation if
you are using that in your textbook, if the
domain is all real numbers except x cannot
equal the value of one then our domain is
going to be from negative infinity to one...then
from one, and this means or and we are not
including the one in our interval notation
thus we are excluding it....from one to infinity.
What do you say the domain, here all real
numbers except x cannot equal one or this
interval notation. Since the domain includes
all real numbers except this value of one
and the numerator and denominator of our rational
function are polynomials, then we are guaranteed
that we are going to be continuous on this
closed interval of [3/2,5/2] Now since we
just checked to make sure that the function
is continuous on this closed interval, now
we need to find f(a) and f(b). I am going
to step out to speed up this video a little
bit. We are going to find f(3/2) and f(5/2).
I am plugging in the lower value of our closed
interval of 3/2. We have f(3/2) and of course
squaring a fraction, multiplying the 2 times
3/2, finding common denominators, and we get
down to a value of 15. So f(3/2) is equal
to 15. f(5/2), not showing that work again
with all of the fractions, comes out to be
35/3 which is approximately 11.667. Ok, so
what the means is...and I will write this
up in a second...we have shown just at least
verbally and I will write it down in a final
statement here...how we checked for the fact
that this function is continuous on this closed
interval, we f(a) is different than f(b),
and our value of k which is 12 is in between
those values. So, we are guaranteed a value
of c, or a c value, where f(c) is going to
be equal to 12. Nananananana... f(x) is continuous
on the closed interval and f(3/2) is greater
than 12 which is greater than f(5/2), so by
the Intermediate Value Theorem there is a
c on the closed interval...this one... such
that f(c) is equal to 12. Now I didn't write
it, or actually I wrote it, I did not read
in the notes how the intermediate value theorem
guarantee a value of c such that f(c) is equal
to k if you meet all of these conditions...but
it does not show you...it does not give it
to you. You have to solve for c. Let's go
ahead and find that. We are looking for a
value of c. We are going to solve the function
such that it is equal to 12. Our function
here is .... what is c because that is part
of the question here... We have got (2x^2+2x)
over (x-1) is equal to 12. We are going to
go ahead and multiply both sides of the equation
by x-1. We are going to get 2x^2 plus 2x is
equal to...and I don't have a ton of space
here so I am going to go ahead and distribute
this...it is 12 times this binomial so 12
times x is 12x and 12 times negative one is
-12. We have of course a quadratic. We also
have single degree terms. We are going to
set this equal to zero by subtracting both
sides by 12x and adding 12 to both sides.
We get 2x^2. 2 minus 12 is negative 10x. Plus,
because we are bring that over with addition...
so 2x^2-10x+12 is equal to zero. Now all of
my coefficients are equal so I can go ahead
and divide both sides of the equation by 2.
We are going to get x squared minus 5x plus
6 is equal to 0. Now this quadratic of course
I am trying to make this easy to do in front
of a camera and we don't need the quadratic
formula to solve this, though that could certainly
be possible. This is factorable. We get x...we
need factors of 6 that add to be negative
five. We have x-3 times x-2 which is equal
to zero. That means that we have x is equal
to 3 and x is equal to 2. So there is actually
two solutions to this equation. The two values
that I can plug into this expression that
would make it equal to 12. I don't want both
of these. I am looking for the c that is in
the closed interval. (what did I mumble?)
Hence here, we know by the intermediate value
theorem there is a c on the closed interval.
That is what we are looking for. So the closed
interval was from 3/2, or 1/5, to 5/2 halves
which is 2.5. While this three is a solution,
this the answer that is in this closed interval.
Thus that is the answer that I am looking
for as a solution to our first example. Next
example!!! Verify the intermediate value theorem
to show that the function has a zero in the
indicated interval. So our function is g(x)
is equal to x cubed minus 3 x squared minus
16x plus 48. The zero, a root, a solution.
A solution to an equation is an x intercept.
So that is what we are looking for when we
are being told to verify or show that the
function has a zero on this closed interval.
We are looking and we want to state and ensure
that with the intermediate value theorem,
that the graph is crossing the x axis somewhere
on the closed interval from negative five
to negative three. That means that we are
looking to basically put in... I wrote g(x)
on purpose. If you are just thinking that
every function should be just called f and
everything is an f(x), and I give you a g(x)
and all of a sudden all of your work started
saying f(x) what is f(x). I didn't give you
f(x), I gave you g(x). Don't change the name
of these functions. That can get you in trouble
on tests. We want to basically look at g(-5),
we want to look at g(-3) and hopefully at
the endpoints of this closed interval we see
a sign change. One of these values giving...I
am not sure which one yet...but one them giving
a negative answer and the other a positive
answer thus showing that we are passing through
the...the graph is going to pass through the
x axis. Now we are not guaranteed that a c
exists such that there is a zero, or that
f(c) is equal to zero unless the function
is continuous. I don't think I really harped
on that enough, how important it is that for
you to remember not just the vague idea of
what the Intermediate Value Theorem says but
that condition of the function being continuous.
Here we have a function which is discontinuous.
It has a non-removable discontinuity. There
is a big gap here and not just a hole which
we would call a removable discontinuity. f(a)
is not equal to f(b). With the closed endpoints
we do have a closed interval. But, the function
is not continuous. I am going to just putting
a k right through the middle of this discontinuity.
A k which is between f(a) and f(b), but there
is no value of c such that f(c)=k. There is
not at least one, there is actually none with
this example. So you have to make sure that
you are going to...when the problem says use
intermediate value theorem...now I am more
blatantly telling you... Hey, you better make
sure that you verify that it applies. You
are going to be first looking at whether or
not the function is continuos. The... I keep
saying function out of habit. For the intermediate
value theorem to hold it does not need to
be a function. The equation or f is continuous.
Well this is a function ok, and we are going
to see here. This is a polynomial and this
polynomial...polynomials are continuous through
everything from negative infinity to positive
infinity. All real numbers. So if polynomials
are continuous everywhere, then certainly
it is going to be continuous on this closed
interval. I have done that check. I will write
it out for you as I reveal my answer here
in a second. I have checked for continuity
on this closed interval. I don't care if it
is continuous anywhere else. I am looking
for a value of c that is guaranteed within
this interval such that f(c) is equal to zero.
Now I want to check g(-5) and g(-3) and hopefully
see a value that is less than zero and a value
which is greater than zero...or in this case
I can just say a sign change. So we have our
arithmetic and our final statement. g(x) is
continuous on the closed interval of negative
from negative five to negative three and g(-5)
is less than zero which is less than g(-3).
By the Intermediate Value Theorem g(x) must
have a value of zero, or have a zero, or there
must be at least one c that exists such that
g(c) is equal to zero on the closed interval
of [-5,-3]. Now I didn't actually ask for
us to find the zero in the problem. But it
never hurts to do a little review and let's
just see here. We have got x^3-3x^2-16x+48
is equal to zero. A lot of third and fourth...a
lot of equations that have a degree or two
are quite difficult to solve. But this one
is going to be factorable by grouping. In
the first two terms we give an x squared from
both of these. In the last two terms, they
can both give up a negative sixteen. By doing
so, now we have two new terms that both have
a factor...if I watched my signs correctly...
Both of these have a factor of x-3, so we
factor out an x-3 and we get minus (x-3)...take
that out and we are left with x squared...
now minus 16. I should have lined this up
so that my equal sign came straight down.
This is the difference of squares. (x-3) times
(x+4) times (x-4) is equal to zero. I set
each of these factors equal to zero to get
solutions of x can be equal to negative four,
three, and positive four. Actually our equation
had three zeros. Now again we are looking
for that c that is guaranteed by the Immediate
Value Theorem in this closed interval. So
the only answer that I concerned with for
this particular question is x is equal to
negative four. Let's take a look at an intermediate
value problem that deals with data that is
given to us in a table. We have to analyze
the table. The table below represents a continuous
function f(x)... It is a function for this
case again... where f(x) is the value of an
investment account. Now investments account.
This is not like a bank account that just
changes whenever you put money in or out of
it, or it just changes at the end of the month
when interest is deposited. Maybe this is
based off the stock market and this is changing
constantly. We have taken one measurement
of the value of the account once a day. Our
dates are the ninth, tenth, eleventh, twelfth,
thirteenth, fourteenth, and fifteenth of some
particular month. Our balances are 7413 dollars
if you will, and 7497, 8105, 7549, 7735, 7390,
and finally 7600 dollars. The question here
is, what is the least number of times this
account was exactly $7500? Justify your answer.
Ok, well it... We are looking for...let's
see here... We are told that the function
that the function is continuous and then we
are looking to see how often it met or equalled
a particular value, a k. Ok, so basically
how many c's are there. How many did this
continuous function... As soon as you read
continuous function, one of the theorems you
need to think about immediately that has the
condition of being continuous is the Intermediate
Value Theorem. As I look across this table
of values, I am going to be looking for, with
this continuous function, anywhere the function
has values that cross that threshold of $7500.
So we have at day 9 7413, day 10 7497, and
if this a continuous...if this function is
constantly changing up and down by the minute
with the stock market, maybe it went above
7500 but these the data points. I can't just
start making up stuff. But I do see that from
the 10th to the 11th our accounts value goes
from being below 7500 to being above. Again
it is continuous, so now...I will write the
formally in a second like I have with my previous
answers, but continuous function basically
f(10) is less than 7500 and f(11) is greater
than so there is at least...and it might have
oscillated within that 24 hour time period...but
we are guaranteed at least one point in time
where the function's value or the accounts
value was $7500. It stays above 7500 until
we get between day 13 and day 14. There is
at least one time period that we are guaranteed...at
least...one within here because f(13) is greater
than 7500 and f(14) is less than 7500. Then
we rise above again the 7500 dollar mark as
we go from day 14 to day 15. So we have another
closed interval where the function goes from
being below 7500 to being above. At least
three from those values changing due to the
Intermediate Value Theorem and the continuity
of the function which is stated. The closed
intervals from [10,11], f(10) again was less
than 7500 which was then less than f(11).
F(13) is greater than 7500 which is then greater
than f(14). Then f(14) is less than 7500 which
is less than f(15). So we went from below
to above, from above to below, and from below
to above. Again, the key word is at least
3 time points or three value of c where the
account values was equal to 7500. I am Mr.
Tarrou. Bam!!! Go Do Your Homework:D
