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Math Help - basic calc two problems! thanks for the help!

  1. #1
    wpski
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    Exclamation basic calc two problems! thanks for the help!

    ok two problems both are totally weird and have no context in my book
    please help me!

    1) The tangent to y=(ax+b)/(sqaureroot of x) at the point where x=1 is 2x-y=1. find a and b

    2) Find a given that the tangent to y=(-2)/((ax+1)^2) at x=0 passe through the point (1,0)


    any help is appreciated
    thank you for all your help!

    ps
    this is my first chapter using calc so like limits and derivatives ect.
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  2. #2
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    Quote Originally Posted by wpski View Post
    ...

    1) The tangent to y=(ax+b)/(sqaureroot of x) at the point where x=1 is 2x-y=1. find a and b

    ...
    Hello,

    you have:

    f(x)=\frac{ax+b}{x^2} = \frac ax + \frac b{x^2}

    and the equation of a tangent line:

    t:\ y=2x-1

    1. Calculate the coordinates of the tangent point for x = 1. Plug in this value into the equation of the tangent line: T(1, 1)
    Plug in these values into the equation of the function because the coordinates of the tangent point must satisfy the equation of the function too:

    f(1) = 1 =\frac{a+b}{1}~\implies~\boxed{a+b=1}~\color{blue}[1]

    2. The slope of the tangent line is m = 2 and is equal to the first derivatie of f at x = 1. Calculate the first derivatie. I took the 2nd version of the term of the function:

    f'(x) = -\frac a{x^2} - \frac{2b}{x^3} . Thus

    f'(1) = -\frac a{1} - \frac{2b}{1}~\implies~\boxed{-a-2b = 2}~ \color{blue}[2]

    3. Solve the system of simultaneous equations for a and b. I've got a = 4 and b = -3
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  3. #3
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    Quote Originally Posted by wpski View Post
    ...

    2) Find a given that the tangent to y=(-2)/((ax+1)^2) at x=0 passe through the point (1,0)

    ...
    Hi,

    you have: f(x) = -\frac{2}{(ax+1)^2} = -2 \cdot (ax+1)^{-2}

    The point T(0, -2) belongs to the graph of f and is the tangent point of a line which passes through the points P(1, 0) and T(0, -2).

    Use the 2-point-formula to calculate the equation of the tangent line:

    \frac{y-0}{x-1} = \frac{-2-0}{0-1} = 2~\implies~y = 2x-2

    Thus the slope of the tangent line is m = 2

    The slope of the tangent line is equal to f'(0)

    f'(x) = (-2) \cdot (-2) \cdot (ax+1)^{-3} \cdot a = \frac{4a}{(ax+1)^3} . That means f'(0) = \frac{4a}{1^3}=4a

    So you get:

    4a = 2~\iff~a=\frac12
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