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Math Help - Basic Calculus HELP!!!!!!!! (:-)

  1. #1
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    Exclamation Basic Calculus HELP!!!!!!!! (:-)

    Find the gradient of the tangeant to the curve:


     <br />
f(x) = x^3/2 - 5x^2 at x = 2

    Find the coordinates of the points at the curve:

    f(x) = (3x - 2)^2 at which the tangeant is parralel to the line y - 6x = 8.

    Find the coordinates of the points on the curve y = x^3 - x^2 + 5x - 3 at which the tangeant is perpindicular to x + 12y = 38

    Find the equation of the curve y = f(x) given f'(x) = 4x^2 - x + 1 and f(0) = 3.
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  2. #2
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    Quote Originally Posted by mibamars View Post
    Find the equation of the curve y = f(x) given f'(x) = 4x^2 - x + 1 and f(0) = 3.

    f'(x) = 4x^2 - x + 1 and f(0) = 3.
    f(x) = (4/3)x^3 - (1/2)x^2 + x + c
    f(0) = 3 = (4/3)0^3 - (1/2)0^2 + 0 + c
    Therefore, c = 3
    f(x) = (4/3)x^3 - (1/2)x^2 + x + 3
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  3. #3
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    Quote Originally Posted by mibamars View Post
    (1) Find the gradient of the tangeant to the curve:

     <br />
f(x) = x^3/2 - 5x^2 at x = 2

    (2) Find the coordinates of the points at the curve:

    f(x) = (3x - 2)^2 at which the tangeant is parralel to the line y - 6x = 8.

    (3) Find the coordinates of the points on the curve y = x^3 - x^2 + 5x - 3 at which the tangeant is perpindicular to x + 12y = 38

    (4) Find the equation of the curve y = f(x) given f'(x) = 4x^2 - x + 1 and f(0) = 3.
    Hello,

    to (1):

    Calculate the first derivative of f: f'(x) = \frac32 x^2 - 10x and plug in x = 2. That means you are calculating f'(2) = -14

    to (2):
    Rearrange the equation of the given line to: y = 6x+8. The slope of the line is m = 6. If the tangent is parallel to this line it's slope must be 6 too. That means you are looking for those values of x that f'(x) = 6.

    f'(x) = 2(3x-2) \cdot 3 = 18x - 12

    18x-12 = 6~\iff~ x = 1 . Plug in this value into the equation of the function. You'll get the coordinates of the tangent point as T(1, 1)

    to (3):

    Rearrange the equation of the line x + 12y = 38~\iff~ y = -\frac1{12} x + \frac{19}6. Therefore the perpendicular direction has the slope m = 12. Thus you are looking for those values of x where f'(x) = 12:

    f'(x) = 3x^2-2x+5

    3x^2-2x+5 = 12~\iff~x = \frac13 \pm \frac13 \cdot \sqrt{22}

    You have to plug in these 2 values for x into the original equation of the function to get the y-coordinate of the tangent point.
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