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Math Help - Horizontal and Vertical Tangent Lines

  1. #1
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    Horizontal and Vertical Tangent Lines

    Question: If they exist, find the horizontal and vertical tangent lines of the function f(x) = x(sqrt(4-x^2))

    I think I found the x values at which the tangent lines are vertical and horizontal, but I'm not sure how to find the actual lines.

    My solution (so far):

    f'x = (4-x^2)^0.5 + (0.5x(4-x^2)^-0.5)*-2x
    = ((4-x^2)^0.5)*((4-x^2)-x^2))
    = (4-2x^2)/sqrt(4-x^2)

    Horizontal tangent line is when numerator=0, therefore x= -sqrt2

    Vertical tangent line is when denominator=0, therefore x= +/- 2

    Anyway, I'm not sure if I'm going about the problem correctly.

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Adrian View Post
    Question: If they exist, find the horizontal and vertical tangent lines of the function f(x) = x(sqrt(4-x^2))

    I think I found the x values at which the tangent lines are vertical and horizontal, but I'm not sure how to find the actual lines.

    My solution (so far):

    f'x = (4-x^2)^0.5 + (0.5x(4-x^2)^-0.5)*-2x
    = ((4-x^2)^0.5)*((4-x^2)-x^2))
    = (4-2x^2)/sqrt(4-x^2)

    Horizontal tangent line is when numerator=0, therefore x= -sqrt2

    Vertical tangent line is when denominator=0, therefore x= +/- 2

    Anyway, I'm not sure if I'm going about the problem correctly.

    Thanks
    recall that vertical lines are of the form: x = c, where c is a constant. so provided your calculations are correct (i did not check them), your vertical tangent lines are x = 2 and x = -2

    horizontal lines are of the form y = c, where c is a constant, so for the x = +/-sqrt(2), just plug those values into the orginal formula to find the corresponding y-coordinate. then your horizontal tangent lines would be y = a and y = b, where a and b are the y-coordinates for x = sqrt(2) and x = -sqrt(2) respectively
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