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Math Help - More function problems...

  1. #1
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    More function problems...

    Find the points on the curve y = 2/3x + 2 where the tangent is parallel to the line y = (-3/2x) - 1

    I know the slope of the tangent is -3/2 as it is parallel to the line y = (-3/2x) - 1.

    Therefore m = -3/2

    So I can express the curve as...

    y = 2(3x - 2)^-1

    Using a combination of the product rule and chain rule I have come up with:

    = -6/(3x - 2)^2 + 1/(3x - 2)

    So...

    -3/2 = -6/(3x - 2)^2 + 1/(3x - 2)

    But that's where I get hung up. Can I express that as the following then solve for x?

    -3/2 = -5/(3x - 2)^2 + (3x - 2)

    Thanks!
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  2. #2
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    Hello, Jools!

    Your derivative is wrong . . .


    Find the points on the curve y \,=\, \frac{2}{3x + 2}
    where the tangent is parallel to the line y \,=\, -\tfrac{3}{2}x - 1

    I know the slope of the tangent is -\tfrac{3}{2}.

    I can express the curve as: . y \:=\:2(3x - 2)^{-1}

    The derivative is: . \frac{dy}{dx} \:=\:-2(3x+2)^{-2}\cdot3 \:=\:\frac{-6}{(3x+2)^2}

    So we have: . \frac{-6}{(3x+2)^2} \;=\;-\frac{3}{2} \quad\Rightarrow\quad -12 \;=\;-3(3x-2)^2

    . . (3x+2)^2 \;=\;4 \quad\Rightarrow\quad 3x + 2 \;=\;\pm2 \quad\Rightarrow\quad 3x \;=\;-2\pm2


    Therefore: . x \;=\;\frac{-2\pm2}{3} \;=\;\begin{Bmatrix}\dfrac{-2+2}{3} &=& 0 \\ \\[-3mm] \dfrac{-2-2}{3} &=& \text{-}\dfrac{4}{3} \end{Bmatrix}

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  3. #3
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    Oh my, it seems to me like you're complicating it.
    The way I see it, you're looking at a spot on the derivative of y where y' = -3/2.
    y = \frac{2}{3x}+2
    y' = \frac{d}{dx}\frac{2}{3x}+\frac{d}{dx}(2)
    Now, the derivative of a constant is 0, so the 2nd term is 0 now.
    Also, remember that 2/3 is also a constant and can thus be taken out of the d/dx.
    y' = -\frac{2}{3x^2}
    -\frac{3}{2} = -\frac{2}{3x^2}
    9x^2 = 4
    I trust you can solve this? (I don't know how to do the plus or minus thing with LaTex)
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  4. #4
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    First I need to point out an error I made here... The curve in question should actually be y = 2/3x - 2, not y = 2/3x + 2. Sorry... my bad. Also, I'm not familiar with dy/dx... How is this being applied, and what does each variable represent? Thanks!
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  5. #5
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    dy/dx means taking the derivative of y with respect to x, or differentiating y with x as the variable of differentiation.
    But remember, though y = 2/3x - 2, it is still y = 2(3x-2)^-1
    y' = -6(3x-2)^-2
    Then plug in y = -3/2
    (3x-2)^2 = 4
    3x - 2 = 2 or 3x - 2 = -2
    x = 4/3 or x = 0
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  6. #6
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    So -6(3x - 2)^-2 I understand, but when we use the product rule, it should equate to F ' (x) = f(x)g ' (x) + f ' (x)g(x) shouldn't it? So what is happening with the second half of the formula?...
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  7. #7
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    And please use parentheses to clarify what you mean!

    You talk about the line "y = (-3/2x) - 1" so that is clearly supposed to be (-3/2)x but you also have "y = 2/3x + 2" which we clearly are supposed to interpret as y= 2/(3x)+ 2 because if we interpret it the same way, as y= (2/3)x+ 2, is a line with slope 2/3 and the two lines are perpendicular.
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  8. #8
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    My apologies. To clarify.... it it 2 / (3x - 2)
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