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Math Help - A Tangent to a Graph

  1. #1
    Newbie pocketasian's Avatar
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    A Tangent to a Graph

    Hello (again), please help me with this problem. I don't really know where to being. Thanks.
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  2. #2
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    start by making f'(x) = 3
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  3. #3
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    Quote Originally Posted by pocketasian View Post
    Hello (again), please help me with this problem. I don't really know where to being. Thanks.
    slope = \frac{d(f(x))}{dx} = 3

    \frac{d(f(x))}{dx} = 16xe^{4x^2} = 3

    solve for x
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  4. #4
    Newbie pocketasian's Avatar
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    Alright, thanks. Since there's an x as an exponent, how should I go about solving for it? Should I take the natural log of each side? And if I do, where do I go from there?
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  5. #5
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    Quote Originally Posted by pocketasian View Post
    Alright, thanks. Since there's an x as an exponent, how should I go about solving for it? Should I take the natural log of each side? And if I do, where do I go from there?
    Use the chain rule to differentiate the given function. If you cannot do this, I suggest you review the technique and appropriate examples from your classnotes and textbook. Note that the correct answer for the derivative has already been given in an earlier post.
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  6. #6
    Newbie pocketasian's Avatar
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    Yes, I know how to solve for the derivative. I just need a push with how to solve for x after I differentiate the function. My algebra is a bit rusty, but I do remember something about taking the natural log of each side to bring the x in the exponent to the front...or something like that.
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  7. #7
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    Dear friends,

    I had thought of this problem for some time. I think this curve does not contain a point such that the tangent is 3. Please see the attachment I have given below. I any of you find the attachment unreadable or any problem with my solution please reply me.

    Hope this helps.
    Attached Thumbnails Attached Thumbnails A Tangent to a Graph-dsc02529.jpg  
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  8. #8
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    Quote Originally Posted by pocketasian View Post
    Yes, I know how to solve for the derivative. I just need a push with how to solve for x after I differentiate the function. My algebra is a bit rusty, but I do remember something about taking the natural log of each side to bring the x in the exponent to the front...or something like that.
    16 x e^{4x^2} = 3 cannot be solved in closed form using algebra. You're undoubtedly expected to use technology to get an approximate solution to a given number of decimal places.
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  9. #9
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    why did you take the derivative twice?

    f'(x) = 3

    16xe^{4x^2} = 3

    x \approx 0.1676
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  10. #10
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    Quote Originally Posted by Sudharaka View Post
    Dear friends,

    I had thought of this problem for some time. I think this curve does not contain a point such that the tangent is 3. Please see the attachment I have given below. I any of you find the attachment unreadable or any problem with my solution please reply me.

    Hope this helps.
    Since f'(0) = 0 and the function increases without bound as x --> +oo it's not difficult to see that a solution to f'(x) = 3 does exist. In fact, the solution lies between x = 0 and x = 1.
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  11. #11
    Newbie pocketasian's Avatar
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    0.168 is the correct answer, which is what skeeter came up with. Thanks everyone for helping me solve this problem. It actually makes better sense now.
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