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    Range: The set of values which a function gives out. ie The positive reals from y=x^2
    Domain: The set of values which a function takes. Positive reals for y=x^2

    Memorising trigonometric identities is up to you. I think most of them are in the formula book, but you do better to memorise them
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    (Original post by SimonM)
    Range: The set of values which a function gives out. ie The positive reals from y=x^2
    Domain: The set of values which a function takes. Positive reals for y=x^2


    Memorising trigonometric identities is up to you. I think most of them are in the formula book, but you do better to memorise them

    so the range for x ^2 is.......

    x \geq 0

    put 0 in the equation and the domain:

    x \geq 0 ??????????????????????????
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    Domain: The possible values of the independent variable
    Range: The possible values of the dependent variable
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    O.o Blimey! You can tell i got an E at Foundation, i have no idea what anybody is on about!
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    Sorry about that, I should have explained myself better.

    If you put the positive reals into y=x^2 you get the positive reals. Therefore positive reals are both domain and range

    If you put the reals into y=x^2 you get the nonnegative reals out. Therefore in this case the reals are domain, nonnegative reals are the range

    Edit

    I'm going to do the job properly this time.

    For those of you with higher knowledge (ie the treatment of functions at university) ignore this. This is aimed to help someone through A levels.

    For A level, a function is defined as three things. A rule, domain and range.

    For example, you could have the rule f(x) = x but that in itself isn't a function without saying what values of x you allow it to take. These could be positive reals, natural numbers, or the numbers 1, 5, and 7.

    If we take a more complicated function, say, y = x^2 and a simple domain, say 1, 2, and 3 then the range will be 1, 4, and 9. With a different domain, say, -1, 1, 2, and 3 you end up with the same range.

    Example

    Find the range of y = x^2 - 3x + 2 with the domain of all the real numbers.

    Solution.

    Clearly y can take any large enough positive real, so it remains to find the lowest number it can take, using calculus we can show this to be when x = 3/2. Therefore the range is the set { x : x >= -1/4 }

    Exercises

    1. Find the range of y = x^2 and y=x^3 over the domain of the real numbers

    2. Find a function which when x = 1, y = 4, x = 2, y = 3, x = 7, y = 3, x = 133, y = 7
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    (Original post by SimonM)
    Sorry about that, I should have explained myself better.

    If you put the positive reals into y=x^2 you get the positive reals. Therefore positive reals are both domain and range

    If you put the reals into y=x^2 you get the nonnegative reals out. Therefore in this case the reals are domain, nonnegative reals are the range

    Edit

    I'm going to do the job properly this time.

    For those of you with higher knowledge (ie the treatment of functions at university) ignore this. This is aimed to help someone through A levels.

    For A level, a function is defined as three things. A rule, domain and range.

    For example, you could have the rule f(x) = x but that in itself isn't a function without saying what values of x you allow it to take. These could be positive reals, natural numbers, or the numbers 1, 5, and 7.

    If we take a more complicated function, say, y = x^2 and a simple domain, say 1, 2, and 3 then the range will be 1, 4, and 9. With a different domain, say, -1, 1, 2, and 3 you end up with the same range.

    Example

    Find the range of y = x^2 - 3x + 2 with the domain of all the real numbers.

    Solution.

    Clearly y can take any large enough positive real, so it remains to find the lowest number it can take, using calculus we can show this to be 3/2. Therefore the range is the set { x : x >= 3/2 }

    Exercises

    1. Find the range of y = x^2 and y=x^3 over the domain of the real numbers

    2. Find a function which when x = 1, y = 4, x = 2, y = 3, x = 7, y = 3, x = 133, y = 7
    1st off how do you get 3/2?

    2nd. after a function when it says for all real numbers or x > 2 ... is that referring to domain or range.
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    Gah, functions, never liked them, you had to remember what words mean. Leave words for english!
    Domains are the numbers for which the function works I think. e.g. x^(1/2). This only works for x > 0 (and x = 0).
    Range is a better word, it makes sense for a start. Its a range of numbers which are given out by a function. Sometimes its easy, f(x) = sin x, sin values are always between 1 and -1, so that'd be the range. If x was bounded though, 0 < x < pi/2 then the range would be 0 <f(x) < 1. The function can get more complex but the idea is that same. The only problem with these two is that its easy to get caught out by them because you forget a certain value of x or something.
    Inverse functions were taught to me this way. f(x) is some function of x, and you want to find the inverse. Call f(x) = y, and rearrange the equation to get x in terms of y. Then change x to f^-1 (x) and y to x. e.g. f(x) = \frac {x + 1}{x - 2}
    y = \frac {x + 1}{x - 2}
    (x-2)y = x + 1
     x(y - 1) -2y = 1
     x = \frac {1+2y}{y - 1}
     f^-^1 (x) = \frac {1+2x}{x - 1}

    And for trigonometric identities I wouldn't worry about 'memorising' them, but knowing of each one in some way would be useful. Its because if a question comes up you need to know which identity to use, not just go to the formula sheet and throw every one at it till the answer comes out. By the time it was exam month I'd already got most of them stuck in my memory. The 1 + tan^2, and 1 + cot^2 ones can be derived in seconds from sin^2 + cos^2 = 1, which is handy because I'm not sure if they're given. The double angle formulas are handy to know because they're not given and it saves time to just know that sin(2x) = 2sin(x)cos(x) and cos(2x) = cos^2(x) - sin^2(x).
    Anyway I hate C3, it holds my lowest mark out of all maths and further maths modules.
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    (Original post by kam_007)
    1st off how do you get 3/2?

    2nd. after a function when it says for all real numbers or x > 2 ... is that referring to domain or range.
    Spotted, 5/2 (Sorry)

    And that is normally the the domain
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    (Original post by SimonM)
    Spotted, 5/2 (Sorry)

    And that is normally the the domain
    I think he meant how do you use calculus to get 3/2(well 5/2 now). What method do you use
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    Well, I actually completed the square (since its quicker) but for most people they find it easier to find the first derivative, set to zero and then check the second derivative.
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    (Original post by SimonM)
    Spotted, 5/2 (Sorry)

    And that is normally the the domain
    Have to say i was a tad lucky.. i did mean how did you get there.

    y = x^2 - 3x + 2

    differentiate it...


    y = 2x - 3

    y = 2x - 3 when y=o

    x = 3/2.
    :mad: :mad: :mad: :mad: :mad: :mad: :mad: im not getting this!!!!! sorry guys.
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    Ignore me, the minimum occurs when x = 3/2. It gives a y value of -1/4

    I've been really stupid today

    Also, write y' = 2x -3 not y = 2x -3
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    (Original post by SimonM)
    Ignore me, the minimum occurs when x = 3/2. It gives a y value of -1/4

    I've been really stupid today
    O soo what ever the equation is. Differentiate it. Equal it to zero. whatever the value of x that gives you the domain??

    And put that value in the equation to get the range??

    How do you if its <, >, less than-equal to etc..........
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    got some other questions:

    1. (x^3) + (3x^2) - 5 = 0

    can be written as:
    x = (root) (5 / (x+3) )

    2. x^3 - x^2 + 8 = 0 has an interval between (-2 and -1)
    Use a suitable oteration forumla to find a approximate to 2 dp.

    x^2 = x^3 + 8
    x = x^2 + 8/x but i don't get a value for x zero???
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    bump-erdyy bump bump.....
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    1) Take 5 over to the right hand side, then take a factor of x^2 on the LHS side.
    2) I'm guessing its following on from the question above and I suggest you take 8 over the RHS and then take a factor of x^2 out and then put x on its own. Then put values in to get the approximation you want. Although I think you've missing some words out in this question.
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    (Original post by insparato)
    1) Take 5 over to the right hand side, then take a factor of x^2 on the LHS side.
    2) I'm guessing its following on from the question above and I suggest you take 8 over the RHS and then take a factor of x^2 out and then put x on its own. Then put values in to get the approximation you want. Although I think you've missing some words out in this question.
    the exact quesiton:
    9a. Show the equation (x^3) - (x^2) + 8 = 0 has a root in the interval (-2, -1)
    b. Use a suitable iteration formula to find an approximation to 2dp for the negative root of the equation (x^3) - (x^2) + 8 = 0.

    edit. they are different questions
 
 
 
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