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    The point (-2,-1) lies on the graph of a function whose derivative f'(x) = 3x² - 12. Find f(x). I don't know exactly what I'm suppose to do with the points. I ended up with:

    f(x) = x³ - 12x

    That looks completely wrong and too simple. Any help would be appreciated.
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    You have started correctly but forgotten the constant of integration

    you should have

    f(x) = x^3 - 12x + c

    Then use your given point to find c
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    You forget the "+ Constant of Integration"

    You are told the point through which the graph passes, so you must now use this to find the exact value of this constant.
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    Remember to add c...
    When integrating you always add c
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    (Original post by daisyclouds)
    The point (-2,-1) lies on the graph of a function whose derivative f'(x) = 3x² - 12. Find f(x). I don't know exactly what I'm suppose to do with the points. I ended up with:

    f(x) = x³ - 12x

    That looks completely wrong and too simple. Any help would be appreciated.
    You've got the right idea, but remember you have a constant when integrating something. So it'll be  f(x) = x^3 - 12x + c

    Then use the values from the point to figure out c.


    EDIT: Beaten to it by a fair few people - Maths forum is on top form tonight!
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    Don't forget to add a constant when you integrate.

    f(x)=x^3-12x+c

    You can use the point on the curve that they gave you to calculate c

    EDIT: ...wow
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    remember when you integrate it becomes x3+12x + C <-- thats the important bit.

    You can then work out C as you have the pints (-2 which is x and -1 which is the f(x))
    f(x)=x^3-12x+c
    put the two coordinates in

    -1=-2^3 +24 + c

    solve the equation to find out c and then put the c into the equation
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    Boy you peeps are slow

    LoL

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    Don't forget to add the consta- oh forget it, the train has left the station, stopped at the end of the line and the driver is taking his lunch break.
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    so I substitute (-2,-1) into the equation:

    -1 = (-2)³ - 12(-2) + c
    -1 = -8 +24 + c
    -1 = 16 + c
    c = -17

    meaning my final equation would be:

    f(x) = x³ - 12x - 17 ?
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    (Original post by TenOfThem)
    when you change that c to -17 yes
    haha yeah just did (edit), my bad :P
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    (Original post by daisyclouds)
    so I substitute (-2,-1) into the equation:

    -1 = (-2)³ - 12(-2) + c
    -1 = -8 +24 + c
    -1 = 16 + c
    c = -17

    meaning my final equation would be:

    f(x) = x³ - 12x - 17 ?
    Looks good to me.
 
 
 
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