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

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    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 /
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    0. Which is obviously a problem
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    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
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    0. So let's have a look at
    what the graph of this
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    function actually looks like.
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    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 -
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    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.
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    But what's actually happening at
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    this point? X equals 0 when the
    arguments is zero. What is going
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    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.
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    Now.
  • 22:47 - 22:52
    If we start off with a value of
    1/2 of one and remember F of X
  • 22:52 - 22:55
    was just equal to one over X.
  • 22:55 - 23:01
    Half of 1 just gives US1, so if
    I get closer to 0 again, let's
  • 23:01 - 23:03
    look at half of 1/2.
  • 23:03 - 23:06
    At 1 / 1/2.
  • 23:06 - 23:11
    This one over 1/2 which just
    gives us 2.
  • 23:11 - 23:15
    So about F of
  • 23:15 - 23:21
    110th. It just gives us 1 / 1
    tenth, which gives us 10.
  • 23:22 - 23:30
    Half of one over 1000 will
    just give us 1 / 1
  • 23:30 - 23:33
    over 1000. Which gives
  • 23:33 - 23:40
    us 1000. What about
    one over 1,000,000, so F of
  • 23:40 - 23:43
    one over a million?
  • 23:43 - 23:48
    It's actually just in the same
    way as before, just going to
  • 23:48 - 23:49
    give us a million.
  • 23:49 - 23:52
    So we can see.
  • 23:53 - 23:57
    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,
  • 24:18 - 24:20
    which is minus one.
  • 24:20 - 24:24
    F of minus 1/2.
  • 24:24 - 24:28
    It's going to be just one
    divided by minus 1/2.
  • 24:28 - 24:30
    Just give his minus 2.
  • 24:31 - 24:34
    Half of minus
  • 24:34 - 24:41
    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.
  • 24:47 - 24:50
    OK so F of minus one over 1000.
  • 24:51 - 24:53
    Will actually give us minus
  • 24:53 - 24:59
    1000. And F of
    minus one over 1,000,000.
  • 24:59 - 25:06
    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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