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Math Help - Tangent

  1. #1
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    Lightbulb Tangent

    find the equation of a line that is tangent to the graph of the given function at the specified point.
    y=x^5-3x^3-5x+2;(1,-5)

    I have an exam coming up on these and I pulled this out of my book to practice...if you could show how to do this one so I can try the rest on my own it would be a big help..Thanks!
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by lisa1984wilson View Post
    find the equation of a line that is tangent to the graph of the given function at the specified point.
    y=x^5-3x^3-5x+2;(1,-5)

    I have an exam coming up on these and I pulled this out of my book to practice...if you could show how to do this one so I can try the rest on my own it would be a big help..Thanks!

    The derivative function will be used to measure slope at the given point. Once we have the slope we can use the point-slope formula to find the equation of the tangent line.

    The derivative is f'(x) = 5x^4  - 9x^2 - 5

    Evaluation at x = 1, we get f'(1) = 5 - 9 - 5 = -9

    so slope is -9 and passes through (1, -5).

    Good luck!
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  3. #3
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    Lightbulb Tangent

    y=2x^4- √x+3/x;(1,4)
    derivative I got is:
    8x^3-1/2-3/x^2
    which then I solved by filling in the x with 1 and got the answer 4.5 slope
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  4. #4
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by lisa1984wilson View Post
    y=2x^4- √x+3/x;(1,4)
    derivative I got is:
    8x^3-1/2-3/x^2
    which then I solved by filling in the x with 1 and got the answer 4.5 slope

    Lisa:

    your derivative seems incorrect. The derivative of \sqrt{x} is \frac{1}{2} x^{\frac{-1}{2}}.
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  5. #5
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    Lightbulb

    radicals are not my thing but I need to get them down...
    I got this as the derivative:
    8x^3-11/4x^2
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  6. #6
    Senior Member apcalculus's Avatar
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    The derivative is:

    8x^3-\frac{1}{2\sqrt{x}} -\frac{3}{x^2}
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