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HELP:how to do this differentiate q

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  • 23-01-2010 7:31pm
    #1
    ayumi
    Registered Users Posts: 676 ✭✭✭


    Im really stuck,I tried it for hours!
    this not on exam papers also its ordinary maths

    the first q is
    a) let f(x)=x^3+2x^2-1
    (x to the power of 3 and 2x to the power of 2)

    i)find f ′(x),the derivative of f(x)

    ii)find the slope of the tangent to the curve y=f(x) at the point where x=-1

    iii)find the two values of x at which the tangents to the curve y=f(x) have a slope of 7

    i really thank anyone who helps me


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    #2
    PunkFreud
    Registered Users Posts: 308 ✭✭PunkFreud


    ayumi wrote: »
    Im really stuck,I tried it for hours!
    this not on exam papers also its ordinary maths

    the first q is
    a) let f(x)=x^3+2x^2-1
    (x to the power of 3 and 2x to the power of 2)

    i)find f ′(x),the derivative of f(x)

    ii)find the slope of the tangent to the curve y=f(x) at the point where x=-1

    iii)find the two values of x at which the tangents to the curve y=f(x) have a slope of 7

    i really thank anyone who helps me

    i) Here you need to get the derivative. So just follow the normal rules:

    f'(x) = 3x^2 + 4x

    ii) Finding the derivative is just another name for finding the slope of something. So it's as simple as subbing in the value of x into the derivative (which we conveniantly found in the last part):

    3(-1)^2 + 4(-1) => 3 - 4 => -1


    iii)Finally, the hardest part, but actually still easy. Remember how in the last part we subbed in -1? Well this time we are trying to find that number, so we let the derivative equal to 7.

    3x^2 + 4x = 7
    3x^2 + 4x -7 = 0 which is a quadratic equation. So lets's solve it
    (3x + 7)(x - 1) = 0
    x = -7/3 ; x = 1

    SOLVED


  • Options
    #3
    ayumi
    Registered Users Posts: 676 ✭✭✭ayumi


    PunkFreud wrote: »
    i) Here you need to get the derivative. So just follow the normal rules:

    f'(x) = 3x^2 + 4x

    ii) Finding the derivative is just another name for finding the slope of something. So it's as simple as subbing in the value of x into the derivative (which we conveniantly found in the last part):

    3(-1)^2 + 4(-1) => 3 - 4 => -1


    iii)Finally, the hardest part, but actually still easy. Remember how in the last part we subbed in -1? Well this time we are trying to find that number, so we let the derivative equal to 7.

    3x^2 + 4x = 7
    3x^2 + 4x -7 = 0 which is a quadratic equation. So lets's solve it
    (3x + 7)(x - 1) = 0
    x = -7/3 ; x = 1

    SOLVED

    thank you for taking the time to do this!:)


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