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    the question is

    the normals to the curve y=√x at the points where x=1 and x=4 meet at P, find the coordinates at P

    ive tried loads of diff things but im jus not gettin the right answer

    so to start of with when x=1 y=1 when x=4 y=2

    and because its the normal the gradients will be the reciprocals(sp?) of the gradients of the tangent so

    dy/dx= -1/2x^-3/2 so grad of tangent when x=1 = -1/2 so grad of normal is 2 and when x=4 grad of tangent is -1/16 so grad of normal is 16

    now what...cause all i tried after this point didnt work

    thanx in advance
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    y = x^(1/2)

    dy/dx = (1/2)x^(-1/2) = 1/(2√x)

    x = 1, y = 1, dy/dx = 1/2
    x = 4, y = 2, dy/dx = 1/4

    Gradient of normal when x=1 ---> -1/(1/2) = -2
    Gradient of normal when x=4 ---> -1/(1/4) = -4

    Therefore y = -2x + c
    Sub x = 1 & y = 1 in:
    1 = -2 + c
    --> c = 3
    ---> y = -2x + 3

    Also, y = -4x + c
    Sub x = 4 & y = 2 in:
    2 = -16 + c
    --> c = 18
    ---> y = -4x + 18

    Equate two normals:
    -2x + 3 = -4x + 18
    2x = 15
    x = 7.5

    Sub x = 7.5 into either of the two normal equations:
    y = -2(7.5) + 3
    y = -15 + 3
    y = 12 -12

    ---> (7.5,12-12)

    This could be wrong, go through my working.

    EDIT: Fixed errors
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    your differentiation seems to be wrong...dy/dx= -1/2x^-3/2 should be dy/dx= +1/2x^-1/2

    this will affect both of your gradients.

    you need the whole equation of the normal , not just the gradient.

    the equation will be y = mx + c where m is the normal eg -2 and c is the intercept.
    to find c use the fact that the point (1,1) is on the line...

    1= -2*1 + c
    c = 3

    so y = -2x + 3

    then you repeat for the second normal.
    once you have both equations you solve them simultaneously.

    and Robert is your mother's brother-in-law
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    (Original post by TomX)
    y = x^(1/2)

    dy/dx = (1/2)x^(-1/2) = 1/(2√x)

    x = 1, y = 1, dy/dx = 1/2
    x = 4, y = 2, dy/dx = 1/4

    Gradient of normal when x=1 ---> -1/(1/2) = -2
    Gradient of normal when x=4 ---> -1/(1/4) = -4

    Therefore y = -2x + c
    Sub x = 1 & y = 1 in:
    1 = -2 + c
    --> c = 3
    ---> y = -2x + 3

    Also, y = -4x + c
    Sub x = 4 & y = 2 in:
    2 = -16 + c
    --> c = 18
    ---> y = -4x + 18

    Equate two normals:
    -2x + 3 = -4x + 18
    2x = 15
    x = 7.5

    Sub x = 7.5 into either of the two normal equations:
    y = -2(7.5) + 3
    y = -15 + 3
    y = 12

    ---> (7.5,12)

    This could be wrong, go through my working.
    thank you so much except its 7.5.-12

    but anyways its not even hard i was jus an idiot and though √x = x^-1/2 when its jus a half and so my dy/dx was completely wrong...cheers tho!
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    can anyone help me with this question?

    f(x) = (c-1/c -x)(4 - 3x^2) where c is a positive constant and x varies over the real numbers.

    i) Show that f(x) has one max and one min
    ii) Show that the difference between these values of f(x) at it turning points is:

    4/9(c+1/c)^3

    What is the least value that the difference in (ii) can have for c > 0
 
 
 
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