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Thread: Review Question

  1. #1
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    Review Question

    Hi
    I have some question i am having trouble with:
    1)If the angle between the lines which passes through the points (-2,3) and (4,0) is:
    A$\displaystyle 2y=x+4$
    B$\displaystyle y=\frac{-x}{2}-2$
    C$\displaystyle 2y+x=4$
    D$\displaystyle y=\frac{x}{2}-2$
    E$\displaystyle 2y-x=4$

    2)Find the derivative of $\displaystyle (4x+\frac{9}{x})^2$ and values of x at which the derivative is zero

    3)Let f be differentiable for all values of x in [2,0]. The graph with equation y=f(x) has a maximum point at (1,3). The equation of the tangent at (1,3) is:
    A$\displaystyle x+3y=0$
    B$\displaystyle x=1$
    C$\displaystyle y=3$
    D$\displaystyle x-3y=0$
    E$\displaystyle 3x+y=0$

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I have some question i am having trouble with:
    1)If the angle between the lines which passes through the points (-2,3) and (4,0) is:

    Have a look at this post...

    http://www.mathhelpforum.com/math-he...-problems.html
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  3. #3
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    Quote Originally Posted by pickslides View Post
    thanks, i know how to do this now
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  4. #4
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    Quote Originally Posted by Paymemoney View Post

    2)Find the derivative of $\displaystyle \left(4x+\frac{9}{x}\right)^2$ and values of x at which the derivative is zero
    Lets say $\displaystyle y = \left(4x+\frac{9}{x}\right)^2$

    Using the chain rule

    $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} $

    Make $\displaystyle u = 4x+\frac{9}{x} $ and $\displaystyle y = u^2$

    Now find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx} $
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  5. #5
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    Quote Originally Posted by pickslides View Post
    Lets say $\displaystyle y = \left(4x+\frac{9}{x}\right)^2$

    Using the chain rule

    $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} $

    Make $\displaystyle u = 4x+\frac{9}{x} $ and $\displaystyle y = u^2$

    Now find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx} $
    This is how far i got until i got stuck.
    $\displaystyle 2(4x+\frac{9}{x})(4+\frac{9}{x^2})$

    $\displaystyle 8(4x+\frac{9}{x})+\frac{9}{x^2}$

    $\displaystyle 32x+\frac{72}{x}-\frac{9}{x^2}$

    $\displaystyle \frac{32x^3}{x^2}+\frac{72x}{x^2}-\frac{9}{x^2}$
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  6. #6
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    $\displaystyle u = 4x+\frac{9}{x} = 4x+9x^{-1}$ and $\displaystyle y = u^2$

    $\displaystyle \frac{dy}{du} = 2u = 2\left(4x+\frac{9}{x}\right)$

    $\displaystyle \frac{du}{dx} = 4-9x^{-2}$

    $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

    $\displaystyle \frac{dy}{dx} = 2\left(4x+\frac{9}{x}\right)\times \left(4-9x^{-2}\right)$

    Now expnad using the F.O.I.L method. Do you know this method?
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  7. #7
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    Quote Originally Posted by pickslides View Post
    $\displaystyle u = 4x+\frac{9}{x} = 4x+9x^{-1}$ and $\displaystyle y = u^2$

    $\displaystyle \frac{dy}{du} = 2u = 2\left(4x+\frac{9}{x}\right)$

    $\displaystyle \frac{du}{dx} = 4-9x^{-2}$

    $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

    $\displaystyle \frac{dy}{dx} = 2\left(4x+\frac{9}{x}\right)\times \left(4-9x^{-2}\right)$

    Now expnad using the F.O.I.L method. Do you know this method?
    oh ok yeh i know what the method is now.
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