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    Differentiation is used to find the gradient of a curve at a particular point (and also obviously to find stationary points - places where the gradient is 0).

    Integration is used to find the area under a curve (or to find the sum of little elements of something, which is what you're doing when you work out the area under a curve).
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    the function is defined for all real values of x by
    f(x)=(x^2+4) (2x-1)

    a) prove that the curve with equation y=f(x) crosses the x-axis at only one point and state the x-coordinates of this point.

    bi) differentiate f(x) with respect to x to obtain f '(x)
    ii) hence show the gradient of the curve y=f(x) is 12 at the point where x=1
    iii) prove that the curve y=f(x) has no stationary point

    c) the curve y=f(x) intersects the line y=x at only one point B
    i) show that the x-coordinates of B satisfies the equation
    2x^3 - x^2 + 7x - 4 = 0
    ii) show that thise equation has a root between 0.56 and 0.57
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    (Original post by elpaw)
    in an AS level maths sense, you use differentiation if you want to find stationary points of a function. you use integration to find the "area under" a function.

    but in the wider mathematical world they have a much wider use

    a lot of physics is solutions of differential equations.
    so it does bring a lot of use in real life then?
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    (Original post by maximusmak)
    the function is defined for all real values of x by
    f(x)=(x^2+4) (2x-1)

    a) prove that the curve with equation y=f(x) crosses the x-axis at only one point and state the x-coordinates of this point.
    This bit is easy. The curve intersects the x-axis at y = 0:

    (2x - 1)(x^2 + 4) = 0

    So, either (x^2 + 4) = 0, or (2x - 1) = 0

    x^2 + 4 has no real solutions, so the only solution is x = 1/2.

    For b:

    Do you know the product rule for differentiation? If not, multiply out the brackets and differentiate:

    f(x) = (x^2 + 4)(2x - 1)

    = 2x^3 - x^2 + 8x - 4

    f'(x) = 6x^2 - 2x + 8

    x = 1: f'(1) = 6.1^2 - 2.1 + 8 = 12

    For a stationary point: f'(x) = 0

    => 6x^2 - 2x + 8 = 0

    Work out the discriminant (b^2 - 4ac): (-2)^2 - 4.6.8

    Clearly, that's less than 0, so there are no real solutions and therefore no stationary points.
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    (Original post by Nylex)
    Differentiation is used to find the gradient of a curve at a particular point (and also obviously to find stationary points - places where the gradient is 0).

    Integration is used to find the area under a curve (or to find the sum of little elements of something, which is what you're doing when you work out the area under a curve).
    thx for the coherent explanation
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    (Original post by maximusmak)
    thx for the coherent explanation
    It's not coherent, elpaw makes more sense, lol.
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    (Original post by maximusmak)
    so it does bring a lot of use in real life then?
    pretty much, yes.
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    (Original post by Nylex)
    This bit is easy. The curve intersects the x-axis at y = 0:

    (2x - 1)(x^2 + 4) = 0

    So, either (x^2 + 4) = 0, or (2x - 1) = 0

    x^2 + 4 has no real solutions, so the only solution is x = 1/2.
    how do you know if if something has no sloutions?
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    (Original post by Nylex)
    It's not coherent, elpaw makes more sense, lol.
    yeah i guess, but his was brief and short and sweet. use was longer and more detailed
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    (Original post by maximusmak)
    how do you know if if something has no sloutions?
    you can't factorise it.

    (or: its determinant is <0)
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    (Original post by elpaw)
    you can't factorise it.

    (or: its determinant is <0)
    oh so it when b^2 - 4ac < 0, i forgot. thx anyway
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    (Original post by maximusmak)
    how do you know if if something has no sloutions?
    As elpaw said, obviously.

    See my post agaim, I've gone through b. for you.
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    (Original post by Nylex)
    This bit is easy. The curve intersects the x-axis at y = 0:

    (2x - 1)(x^2 + 4) = 0

    So, either (x^2 + 4) = 0, or (2x - 1) = 0

    x^2 + 4 has no real solutions, so the only solution is x = 1/2.

    For b:

    Do you know the product rule for differentiation? If not, multiply out the brackets and differentiate:

    f(x) = (x^2 + 4)(2x - 1)

    = 2x^3 - x^2 + 8x - 4

    f'(x) = 6x^2 - 2x + 8

    x = 1: f'(1) = 6.1^2 - 2.1 + 8 = 12

    For a stationary point: f'(x) = 0

    => 6x^2 - 2x + 8 = 0

    Work out the discriminant (b^2 - 4ac): (-2)^2 - 4.6.8

    Clearly, that's less than 0, so there are no real solutions and therefore no stationary points.
    so for all stationary point they have to equal 0 and you have to use the disciminant to work out if its got roots, right?
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    can some 1 remind me of all the sin and cos and tan rules if you want tp substitute them with each other, and if you want to substitute them with something else (i.e. i think you can substite sinx with x for some reason or another)
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    (Original post by maximusmak)
    can some 1 remind me of all the sin and cos and tan rules if you want tp substitute them with each other, and if you want to substitute them with something else (i.e. i think you can substite sinx with x for some reason or another)
    sin(x) ~= x for small x in radians
    cos(x) ~= (1-x²/2) for small x in radians
    tan(x) ~= x for small x in radians

    (by "small", x << 1)
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    (Original post by maximusmak)
    can some 1 remind me of all the sin and cos and tan rules if you want tp substitute them with each other, and if you want to substitute them with something else (i.e. i think you can substite sinx with x for some reason or another)
    tanx = sinx / cosx
    sin^2x + cos^2x = 1

    Remember, in mathematics, "sin^2x" means (sinx)^2.

    Just to make sure, as others have been confused about this.

    eg) (sin50)^2 + (cos50)^2 = 1
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    (Original post by elpaw)
    sin(x) ~= x for small x in radians
    cos(x) ~= (1-x²/2) for small x in radians
    tan(x) ~= x for small x in radians

    (by "small", x << 1)
    thx, so can you give me an example where i can substitute sin(x) with x (when x is a small value) and an example when you cant subsitute sin(x) with x.
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    (Original post by maximusmak)
    so for all stationary point they have to equal 0 and you have to use the disciminant to work out if its got roots, right?
    For all stationary points, dy/dx or f'(x) = 0, yeah.

    You needed to use the discriminant to work out if it had real roots. If so, there were stationary points (b^2 - 4ac = 0 or > 0). If the discriminant was < 0, it'd have imaginary roots and therefore no stationary points.
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    (Original post by Renaissance)
    tanx = sinx / cosx
    sin^2x + cos^2x = 1

    Remember, in mathematics, "sin^2x" means (sinx)^2.

    Just to make sure, as others have been confused about this.

    eg) (sin50)^2 + (cos50)^2 = 1

    dont get the former,can you explain the e.g
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    (Original post by Nylex)
    For all stationary points, dy/dx or f'(x) = 0, yeah.

    You needed to use the discriminant to work out if it had real roots. If so, there were stationary points (b^2 - 4ac = 0 or > 0). If the discriminant was < 0, it'd have imaginary roots and therefore no stationary points.
    thx, just wanted to get that clarified
 
 
 
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