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Math Help - tangent to the function #3

  1. #1
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    tangent to the function #3

    1. Find the values of x so that the tangent to the function y=3/(x^1/3) is parallel to the line x+16y+3=0?

    2. Find the point on the parabola y=-x^2+3x+4 where the slope of the tangent is 5
    Last edited by mr fantastic; August 12th 2009 at 04:08 PM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    1. Find the values of x so that the tangent to the function y=3/(x^1/3) is parallel to the line x+16y+3=0?

    find the derivative of y and set it equal to the slope of the given line, then solve for x.

    2. Find the point on the parabola y=-x^2+3x+4 where the slope of the tangent is 5

    find the derivative of y and set it equal to 5, then solve for x. don't forget that the question asks for a point on the original curve.
    ...
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by skeske1234 View Post
    1. Find the values of x so that the tangent to the function y=3/(x^1/3) is parallel to the line x+16y+3=0?

    2. Find the point on the parabola y=-x^2+3x+4 where the slope of the tangent is 5
    HELLO:There is the solution
    1 ) let : f(x) = 3x^{ - \frac{1}{3}} <br />
    The dirivate is :
    \begin{array}{l}<br />
f'(x) = 3\left( { - \frac{1}{3}} \right)x^{ - \frac{1}{3} - 1} \\ <br />
f'(x) = - \frac{1}{{x^{\frac{4}{3}} }} \\ <br />
\end{array}<br />
    But :

    x + 16y + 3 = 0 \Leftrightarrow y = - \frac{1}{{16}}x - \frac{3}{{16}}
    The tangent is parallel for this line the are same slope - \frac{1}{{16}}<br />
    solve : f'(x) = - \frac{1}{{16}}
     - \frac{1}{{x^{\frac{4}{3}} }} = - \frac{1}{{16}} \Leftrightarrow x^{\frac{4}{3}} = 16
    I"have :
    \left( {x^{\frac{1}{3}} } \right)^4 = 2^4 \Leftrightarrow x^{\frac{1}{3}} = 2 \Leftrightarrow x = 2^3
    CALCULATE :
    f(2^3 ) = 3\left( {2^3 } \right)^{ - \frac{1}{3}} = \frac{3}{2}
    The point is :
    A\left( {8,\frac{3}{2}} \right)
    LOOK HERE THE GRAPHE
    Attached Thumbnails Attached Thumbnails tangent to the function #3-cube.jpg  
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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by skeske1234 View Post
    1. Find the values of x so that the tangent to the function y=3/(x^1/3) is parallel to the line x+16y+3=0?

    2. Find the point on the parabola y=-x^2+3x+4 where the slope of the tangent is 5
    Hello same solution in first question

    \begin{array}{l}<br />
g(x) = - x^2 + 3x + 4 \\ <br />
g'(x) = - 2x + 3 \\ <br />
\end{array}
    Solve this equation :

    g'(x) = 5 \Leftrightarrow - 2x + 3 = 5 \Leftrightarrow x = 5
    Calculate :

    g(5) = - \left( { - 1} \right)^2 + 3\left( { - 1} \right) + 4 = 0

    the point is

    A( - 1,0)
    LOOK HERE THE GRAPHE
    Attached Thumbnails Attached Thumbnails tangent to the function #3-pra.jpg  
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