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Math Help - horizontal tangent line

  1. #1
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    Exclamation horizontal tangent line

    Will you please find all values of x for the given function where the tangent line is horizontal?

    f(x) = x / (x^2 + 1)^3
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  2. #2
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    Two things: how do you find the slope of the tangent lines in general for f(x)? If a line is horizontal, what is its slope? How can you use these two things together to solve?

    If that doesn't help I'll get you started.
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  3. #3
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    Thank you for replying me
    If slope is tangent, that means f'(x) = 0. And its slope is constant.
    Am I right?
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  4. #4
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    Quote Originally Posted by Matho View Post
    Thank you for replying me
    If slope is tangent, that means f'(x) = 0. And its slope is constant.
    Am I right?
    Not quite. Well you might be right and I just am interpreting your post wrong, but I'll post my comment anyway. Disregard if you know this already.

    The slope of a line that is tangent to a certain point on f(x) is different at different points because f(x) increases and decreases at different rates. Taking f'(a) will tell you the slope of a line that is tangent to the point (a, f(a)) on the graph of f(x), but the slope of the point (a+1) is going to be different.

    You are right though that you need to find when f'(x)=0 and solve for x.
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  5. #5
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    Hi Jameson,

    I got f'(x) = (x^6 - x^5 + 3x^4 - 2x^3 + 3x^2 - x + 1) / (x^2 + 1)^5
    However, I don't know how to get f'(x) = 0 from this...
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