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Math Help - linear approximation

  1. #1
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    Apr 2010
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    linear approximation

    Find the value of the derivative for the implicit equation 2x + xy^4 = 3y at the point where y = 2. Give your answer accurate to 3 decimal places. I got -.621 but was marked incorrect. I got dy/dx = (2+y^4)/(3 - 4y^3) then plugged in y = 2. Not sure what I did wrong here.

    For the function y = 3 x^2 - 4 find the point c between -1 and 2 which satisfies the MEAN VALUE THEOREM. GIVE YOUR ANSWER ACCURATE TO TWO DECIMAL PLACES.
    Ok. I did f'(c) = (f(b) - f(a)) / (b - a); Then I got f'(c) = 23 + 1 / 3 + 1 or f(c) = 6; Then I can see that f'(x) = 6x and f'(6) = 1.00; So the answer I got is 1.00 What did I miss on this one?

    Use a linear approximation and the point a = 0 to calculate the approximate value for the function
    y = 3 Arctan(4x) at x = 2 Use RADIANS and give the answer accurate to 3 decimal places. Answer I got is 687.549; I simply found the derivative and put in the value of 3. How do I do this type of problem and what would the correct answer be? Thanks.
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  2. #2
    Senior Member
    Joined
    Feb 2010
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    422
    The denominator should be 3-4xy^3.

    f'(c) = (12-3)/3=3.

    Find the tangent line at x=0. Then put x=2 into that.
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