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## www.mathcentre.ac.uk/.../Introduction%20to%20functions.mp4

• 0:01 - 0:03
What is a function?
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Now, one definition of a
function is that a function is a
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rule that Maps 1 unique number
to another unique number. So In
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other words, If I start off with
a number and I apply my
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function, I finish up with
another unique number. So for
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example, let's suppose that my
function added three to any
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number I could start off with a
#2. I apply my function and I
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finish up with the number 5.
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Start off with the number 8 and
I apply my function and I finish
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up with the number 11.
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And if I start off with a number
X and I apply my function, I
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finish up with the number X +3
and we can show this
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mathematically by writing F of X
equals X plus three, where X is
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our inputs, which we often
called the arguments of the
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function and X +3 is our output.
Now suppose I had an argument of
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two, I could write down F of two
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equals. 2 + 3, which gives us an
output of five.
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Suppose I had an argument of
eight. I could write down F of
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eight equals 8 + 3, which gives
me an output of 11.
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And suppose I had an argument of
minus six. I could write F of
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minus 6. Equals minus 6 + 3,
which would give me minus three
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for my output.
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And if I had an argument of zed,
I could write down F of said
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equals zed plus 3.
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And likewise, if I had an
argument of X squared, I could
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write down F of X squared equals
X squared +3.
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Now it is with me pointing out
here that's a lower first sight.
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It might appear that we can
choose any value for argument.
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That's not always the case, as
we shall see later, but because
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we do have some choice on what
number we can pick the argument
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to be, this is sometimes called
the independent variable.
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And because output depends on
our choice of arguments, the
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output is sometimes called the
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dependent variable. Now let's
have a look at an example
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F of X equals 3 X
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squared. Minus 4.
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Now, as we said before, X is our
input which we call the argument
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and that is the independent
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variable. And our output which
is 3 X squared minus four is
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our dependent variable.
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Now we can choose different
values for arguments, which is
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often a good place to start when
we get function like this. So F
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of zero will give us 3 * 0
squared takeaway 4, which is 0 -
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4, which is just minus 4.
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If we start off with an argument
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of one. We get F of one
equals 3 * 1 squared takeaway 4,
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which is just three takeaway
four which gives us minus one.
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If we have an argument of two F
of two equals 3 * 2 squared
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takeaway, four switches 3 * 4,
which is 12 takeaway. Four gives
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us 8. And as I said before,
there's no reason why we can't
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include and negative arguments.
So if I put F of minus one in.
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The three times minus one
squared takeaway 4, which is 3 *
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1 gives us three takeaway. Four
gives us minus one.
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And F of minus 2.
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Would give us three times minus
2 squared takeaway 4 which is 3
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* 4 gives us 12 and take away
four will give us 8.
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So what are we going to do with
these results? Well, one thing
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we can do is put them into a
table to help us plot the graph
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of the function. So if we do a
table of X&F of X.
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The values were chosen for RX
column, which is. Our arguments
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are minus 1 - 2 zero one and
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two. So minus 2
- 1 zero 12.
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And the corresponding outputs
are 8 - 1.
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Minus 4 - 1 and
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8. And we can use this as I
said to help us plot the
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graph of the function.
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So just to copy the table down
again. So we've got handy.
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We have minus 2
- 1 zero 12.
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8 - 1 - 4
- 1 and 8.
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So we plot our graph.
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Of F of X on the vertical
axis, so output and arguments.
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On the horizontal axis.
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So we've got
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one too. Minus 1 -
2.
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And we've got.
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Minus 1 - 2 - 3 - 4.
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I'm going off.
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Up to 81234567 and eight. So
our first point is minus 2
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eight so we can put the
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appear. Our second point, minus
1 - 1 should appear down here.
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0 - 4.
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Here one and minus one.
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Yeah, but up on two and eight
which will appear over here
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and we can see we can draw a
smooth curve through these
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points, which will be the
graph of the function.
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OK, now why are we drawing a
graph of a function? Because
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this is quite useful to us
because we can now read off the
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output of a function for any
given arguments straight off the
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graph without the need to do any
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calculations. So for example, if
we look at two and arguments of
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two, we know that's going to
give us 8 before I do work that
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out. But if we looked and we
wanted to figure out.
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But the output would be when the
argument was one point 5. If we
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follow our lineup and across you
can see that that gives us a
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value between 2:00 and 3:00 for
the output, and if we substitute
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in 1.5 back into our original
expression for the function, you
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can see actually gives us an
exact value of 2.75.
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Now earlier on when I discussed
uniqueness, I said that a unique
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inputs had to give us a unique
output and by that what we mean
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is that for any given argument
we should get only one output.
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One of the benefits of having a
graph of a function is that we
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can check this using our ruler.
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If we line our ruler up
vertically and we move it left
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and right across the graph.
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We can make sure that the rule I
only have across is the graph
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wants at any point.
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And as we can see, that's
clearly the case in this
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example. And when that happens,
the graph is a valid function.
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Now, if we had the example.
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F of X.
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Equals root X.
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A good place to start is always
to substitute in some values for
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the arguments, so F of 0.
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This gives us the square root of
0, which is 0.
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Half of one's own arguments of
one will give us plus or minus
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one for the square root.
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F of two.
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Will give us plus or minus
1.4 just to one decimal place
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there. Half of 3.
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Which gives us the square root
of 3 gives us plus or minus 1.7.
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An F4. Will give us
square root of 4 which is just
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plus or minus 2.
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Now. If we try
to put in any negative
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arguments here you can see
that we're going to run
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into trouble because we
have to try and calculate
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the square root of a
negative number and we'll
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come back to this problem
in a second. But for now,
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let's plot the points that
we've got so far.
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So if we take out F of X
axis vertical again.
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And our arguments access X
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horizontal. We've
got
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1234.
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And there are vertical axis we
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have minus one. Minus 2.
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Plus one. Plus two points.
We've got zero and zero.
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One on plus one.
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And also 1A minus one.
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We've got two and
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positive 1.4. So round
about that and also to negative
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1.4. We've got three and
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positive 1.7. And we've
got three and negative 1.7.
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And finally we have four and +2.
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And four and negative 2.
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OK, and we've got enough points
here that we can draw a smooth
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curve through these points.
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At something it looks like.
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This.
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OK. Now.
As usual, we will apply our
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ruler test to make sure that the
function is valid and you can
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see straight away that when we
line up all the vertically and
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move it across for any given
positive arguments, we're
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getting two outputs. So
obviously we need to do
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something about this to make the
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function valid. One way to get
around this problem is by
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defining route X to take only
positive values or 0.
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This is sometimes called the
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positive square root. So in
effect we lose the bottom
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negative half of this graph.
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And obviously we also have the
issue of the negative arguments,
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and since we can't take the
square root of a negative
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number, we also have to exclude
these from the X axis. Now when
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we start talking about these
kind of restrictions, it's
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important that we use the right
kind of mathematical language.
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So the set of possible inputs is
what we call the domain, and the
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set of possible outputs is what
we call the range.
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So in this case, when we've got
RF of X equals the square root
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of X. We need to restrict our
domain to be X is more than or
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equal to 0, 'cause we only
wanted the positive values and
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zero. But we also notice that
now because we've got rid of the
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bottom half of the graph.
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The only part of the range which
are included are also more than
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or equal to 0. So range is
defined by F of X more than or
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equal to 0. So now we have a
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valid function. So what will do
now is just look at a couple
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more examples to pull together
everything that we've done so
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far and will start with this
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one. Let's look at the function
F of X equals 2 X squared
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minus three X +5.
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No, as usual, a good place to
start when you get a function is
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to substitute in some values for
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the arguments. So let's start
with that. So now arguments of 0
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would give us F of 0, which is 2
* 0 squared.
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Minus 3 * 0 +
5 which is just zero
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takeaway 0 + 5.
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So we get now pose A5 that if we
had an argument of one.
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We get 2 * 1
squared takeaway 3 * 1
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+ 5. Which gives us 2
* 1 here, which is 2.
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Take away 3 * 1 which is take
away three and plus five. So two
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takeaway three is minus 1 + 5
gives us 4.
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OK, we look at an
argument of two.
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We get 2 * 2 squared.
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Take away 3 * 2.
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I'm plus five which gives us 2 *
4, which is 8 and take away 3 *
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2 which is take away 6 and then
forget our plus five at the end.
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So eight takeaway six is 2.
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+5 gives us 7.
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OK. If we look at an argument of
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three. Half of three gives us
2 * 3 squared.
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Take away 3 * 3.
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And +5.
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So this is 2 * 9 here
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18. Take away 3 * 3 which is
take away nine and +5.
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So 18 takeaway 9 is 9 + 5
gives us 14.
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And last but not least, we can
also include a negative
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arguments, so we'll put negative
arguments of minus one. So F of
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minus one gives us two times
minus 1 squared.
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Take away three times minus one.
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And of course, our +5.
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So two times minus one squared
just gives us 2.
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Takeaway minus sorry takeaway
three times negative one which
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just gives us a plus 3.
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And then we've got a +5.
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2 + 3 + 5 just gives us 10.
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And what we can do is as before,
just put this into a table to
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make it nice and easy to make a
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graph of the function. So we put
it into a table of X.
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And F of X.
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For arguments we
had minus 10123.
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For Outputs, we had 10 five.
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4, Seven and
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14. OK, so let's see the graph
of this function then.
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You start off with our.
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F of X on the vertical axis as
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before. An argument.
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X on the horizontal axis.
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And we've got over 2 - 1, The
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One. 2. And
three, and on the vertical axis.
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Go to 15.
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OK, so our first point to plot
is minus one and 10 which will
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give us something there zero and
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five. There's a point here. One
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and four. It gives the points
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here. Two and Seven.
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Should be around here and
three and 14 which will be
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about here so we can see the
kind of shape that this
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function is starting to take
here. And we can draw in the
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graph.
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What we want is to say that
every input gives us only one
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single output, so we can get our
ruler again and just quickly
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check by going along and we can
see that as we go along. Our
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rule across is the curve once
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and once only. Which means that
this function is valid.
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However, an interesting point to
note is this point here. The
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minimum point which actually
occurs when X is North .75.
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So with X value of North .75
are outputs can take a minimum
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value of 3.875.
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So this means when we look at
our domain and range, we need to
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make no restrictions on the
domain because our function was
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valid. However. Our range has
a minimum of 3.875, so we write
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this. As F of X equals.
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Two X squared minus three X +5.
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And the range F of X has
always been more than or equal
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to 3.875. So for
the next example.
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What would happen if we had a
function F defined by?
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F of X equals
one over X.
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Well, that's always the first
stage is to substitute in some
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values for the arguments.
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So for F of one.
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The argument is one is 1 / 1
just gives US1.
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For Port F of two.
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We just get one half.
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F of three gives us 1/3.
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And F4 will
give us 1/4.
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And as before, we can also look
at some negative arguments.
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So if I look at F of minus one.
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Skip 1 divided by minus one,
which is just minus one.
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F of minus 2.
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Is 1 divided by minus two, which
just gives us minus 1/2?
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F of minus three. Same thing
will give us minus 1/3.
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And F of minus 4.
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Will give us minus 1/4?
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Now if we look at F of 0.
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We have 1 /
• 20:10 - 20:13
0. Which is obviously a problem
• 20:13 - 20:17
for us. Because of this
problem, we have to restrict
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our domain so that it does not
include the arguments X equals
• 20:22 - 20:26
0. So let's have a look at
what the graph of this
• 20:26 - 20:28
function actually looks like.
• 20:29 - 20:33
So let's be 4F of X and are
vertical axis for the Outputs.
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An argument sax on
the horizontal axis.
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OK, so we've gone
over here 1234.
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And minus 1 - 2 -
3 - 4 over here.
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As we go.
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All the web to one down to minus
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one. Has it as well? So
we've got one and one.
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We've got 2
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1/2. 3
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1/3. 4
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one quarter.
We've got minus 1 - 1.
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Minus 2 -
• 21:27 - 21:34
1/2. Minus
3 - 1/3.
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A minus four and minus 1/4.
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OK, and obviously we've excluded
0 from our domain. As we said
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before. So if we join these
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points up. And a smooth curve.
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Get something that
looks like this.
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Now, obviously we've excluded X
equals 0 from our domain, but
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it's also worth noticing here.
Thought there's nothing at the
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output of F of X equals 0, so
that also ends up being excluded
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from the range.
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So we actually end up with F of
X equals one over X.
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And we've got X not equal to 0
from the domain.
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And also. In the range F of X
never equals 0 either.
• 22:29 - 22:31
But what's actually happening at
• 22:31 - 22:35
this point? X equals 0 when the
arguments is zero. What is going
• 22:35 - 22:40
on? Well, let's have a look and
will start off by having a look
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what happens as we get closer
and closer to 0.
• 22:44 - 22:46
Now.
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If we start off with a value of
1/2 of one and remember F of X
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was just equal to one over X.
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Half of 1 just gives US1, so if
I get closer to 0 again, let's
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look at half of 1/2.
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At 1 / 1/2.
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This one over 1/2 which just
gives us 2.
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So about F of
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110th. It just gives us 1 / 1
tenth, which gives us 10.
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Half of one over 1000 will
just give us 1 / 1
• 23:30 - 23:33
over 1000. Which gives
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us 1000. What about
one over 1,000,000, so F of
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one over a million?
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It's actually just in the same
way as before, just going to
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give us a million.
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So we can see.
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The US we get closer and closer
to zero from the right hand side
• 23:57 - 24:00
as we saw on our graph before.
• 24:00 - 24:03
We're getting closer and closer
to positive Infinity to the
• 24:03 - 24:05
graph goes off to positive
• 24:05 - 24:09
Infinity that. What happens when
we approach zero from the left
• 24:09 - 24:11
hand side? Well, let's have a
• 24:11 - 24:18
look. This is minus one, just
gives us 1 divided by minus one,
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which is minus one.
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F of minus 1/2.
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It's going to be just one
divided by minus 1/2.
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Just give his minus 2.
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Half of minus
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110th. Is 1 divided
by minus 110th?
• 24:41 - 24:47
It just gives us minus 10 and we
can kind of see a pattern here.
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OK so F of minus one over 1000.
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Will actually give us minus
• 24:53 - 24:59
1000. And F of
minus one over 1,000,000.
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Actually gives
us minus
• 25:06 - 25:11
1,000,000. So you can see that
as we approach zero from the
• 25:11 - 25:13
left or outputs approaches
• 25:13 - 25:18
negative Infinity. And as we
approach zero from the right
• 25:18 - 25:22
hand side are output approaches
positive Infinity, and these are
• 25:22 - 25:25
very different things. OK, for
the last example.
• 25:26 - 25:30
I just like to look at the
function F defined by.
• 25:30 - 25:38
F of X equals one over
X minus two all squared.
• 25:38 - 25:42
So as always with the examples
we've done, it's worthwhile
• 25:42 - 25:45
started off by looking at some
different values for the
• 25:45 - 25:51
arguments. So we start off with
an argument of minus two, so
• 25:51 - 25:57
half of minus two gives us one
over minus 2 - 2 squared.
• 25:57 - 26:04
Which gives us one over minus 4
squared, which is one over 16.
• 26:04 - 26:07
F of minus one.
• 26:07 - 26:15
Will give us one over minus 1
- 2 all squared, which gives us
• 26:15 - 26:20
one over minus 3 squared which
is one over 9.
• 26:20 - 26:26
Now, FO arguments of zero will
give us one over 0 - 2
• 26:26 - 26:33
all squared, which is one over
minus 2 squared, which works out
• 26:33 - 26:34
as one quarter.
• 26:35 - 26:42
And F of one will give
us one over 1 - 2
• 26:42 - 26:47
squared. Which is just one over
minus one squared, which gives
• 26:47 - 26:48
us just one.
• 26:49 - 26:56
OK. Half
of two gives us one over.
• 26:57 - 27:02
2 - 2 or squared, which gives us
one over 0 which presents us
• 27:02 - 27:06
with exactly the same problem we
had in the previous example when
• 27:06 - 27:12
we had one over X and so we have
to exclude X equals 2 from the
• 27:12 - 27:19
domain. Half of three gives us
one over 3 - 2 all squared.
• 27:20 - 27:25
Which is one over 1 squared is
just gives US1 again.
• 27:25 - 27:32
After four gives us one over 4
- 2 or squared, which gives us
• 27:32 - 27:34
one over 4.
• 27:34 - 27:42
After 5.
Gives us one over 5 - 2 or
• 27:42 - 27:49
squared which gives us one over
3 squared which is 1 ninth and
• 27:49 - 27:56
finally F of six gives us one
over 6 - 2 all squared which is
• 27:56 - 28:01
one over 4 squared which works
out as one over 16.
• 28:02 - 28:07
Now if we want to plot the graph
of this function will probably
• 28:07 - 28:09
need to put this into a table
• 28:09 - 28:16
first. So as usual, do our
table of X&F of X.
• 28:16 - 28:23
OK, and we went from minus 2
- 1 zero all the way.
• 28:24 - 28:26
Up to and arguments of sex.
• 28:27 - 28:29
And the values we got for the
• 28:29 - 28:32
Outputs. One over
• 28:32 - 28:38
16. One 9th,
one quarter and
• 28:38 - 28:44
1:01. One quarter,
1 ninth and
• 28:44 - 28:47
one over 16.
• 28:48 - 28:52
So we plot that onto.
• 28:52 - 28:54
The graph as before.
• 28:57 - 29:01
So we have arguments going along
the horizontal axis.
• 29:01 - 29:04
And Outputs going along
the vertical axis.
• 29:05 - 29:12
We've gone from minus
1 - 2 over
• 29:12 - 29:16
there 123456 along this
• 29:16 - 29:19
way. And then going off, we've
• 29:19 - 29:25
gone too. One up here, so put in
a few of the marks 1/2.
• 29:26 - 29:27
It's put in 1/4 that.
• 29:28 - 29:35
Put in 3/4. OK, so we've got
minus two and 116th, which is
• 29:35 - 29:38
going to come in down here.
• 29:38 - 29:41
Minus one and one 9th.
• 29:41 - 29:47
For coming over here zero
and one quarter.
• 29:47 - 29:50
Over here. 1 on one.
• 29:51 - 29:52
Right, the way up here?
• 29:53 - 29:57
To an. Obviously this was the
divide by zero, so we couldn't
• 29:57 - 30:01
do anything with that. We've
excluded, uh, from our domain.
• 30:02 - 30:03
Three and one.
• 30:04 - 30:07
Pay up. Four and one quarter.
• 30:09 - 30:17
I'm here. Five and one,
9th and six and 116th.
• 30:17 - 30:22
Because we've excluded X equals
2 from our domain.
• 30:23 - 30:26
Put dotted line there,
so that's an asymptotes.
• 30:28 - 30:30
And we can draw our curve.
• 30:32 - 30:33
Up through the points.
• 30:34 - 30:37
On this side.
• 30:38 - 30:42
And we can see differently to
the other example where F of X
• 30:42 - 30:44
is one over X this time.
• 30:45 - 30:50
As we get approach to from both
the left and from the right,
• 30:50 - 30:54
both of the outputs are heading
towards positive Infinity, so a
• 30:54 - 30:58
little bit different, and also
because we've excluded X equals
• 30:58 - 31:00
2 from the domain of function is
• 31:00 - 31:05
now valid. But most also notice
that our range is never zero,
• 31:05 - 31:07
and it's also never negative.
• 31:08 - 31:11
So to write this out properly.
• 31:11 - 31:16
Our function F of X equals one
over X minus two all squared.
• 31:18 - 31:19
And we said.
• 31:19 - 31:21
They are domain is restricted so
• 31:21 - 31:27
it doesn't include two. And our
range is always more than 0.
• 31:28 - 31:31
OK, so let's just recap on what
we've done in this unit.
• 31:32 - 31:34
So firstly, the definition of a
• 31:34 - 31:39
function. And that was that. A
function is a rule that Maps are
• 31:39 - 31:43
unique number X to another
unique number F of X.
• 31:44 - 31:48
Secondly, was the idea that
an argument is exactly the
• 31:48 - 31:50
same as an input.
• 31:52 - 31:56
Thirdly, we looked at the idea
of independent and dependent
• 31:56 - 32:00
variables. And we said that
the input axe was the
• 32:00 - 32:04
independent variable and the
output was the dependent
• 32:04 - 32:04
variable.
• 32:05 - 32:10
4th, we looked at the domain and
we said that the domain was the
• 32:10 - 32:12
set of possible inputs.
• 32:13 - 32:17
And finally we looked at the
range and we said that the range
• 32:17 - 32:18
was the set of possible outputs.
• 32:20 - 32:22
So now you know how to define a
• 32:22 - 32:25
function. And how to find
the outputs of a function
• 32:25 - 32:26
for a given argument?
Title:
www.mathcentre.ac.uk/.../Introduction%20to%20functions.mp4
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