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    the function f(x) is defined by
    f(x) = (2 + sin2x)/(2 + cosx)
    Ignoring terms in x^3 and higher powers of x, obtain a quadratic approximation to the function f(x) for small values of x.
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    (Original post by islandguy)
    the function f(x) is defined by
    f(x) = (2 + sin2x)/(2 + cosx)
    Ignoring terms in x^3 and higher powers of x, obtain a quadratic approximation to the function f(x) for small values of x.
    use maclaurin's series after simplification

    (Original post by islandguy)
    the function f(x) is defined by
    f(x) = (2 + sin2x)/(2 + cosx)
    Ignoring terms in x^3 and higher powers of x, obtain a quadratic approximation to the function f(x) for small values of x.
    (for small x) sin2x is approx 2x, cosx is approx 1-(x^2)/2 (mclaurin)

    therfore

    f(x)=(2 + 2x)(3 - (x^2)/2) = 6 - x^2 +6x - (x^3)

    = 6 + 6x - x^2 (approx ignoring x^3 and higher)

    (Original post by s-man)
    (for small x) sin2x is approx 2x, cosx is approx 1-(x^2)/2 (mclaurin)

    therfore

    f(x)=(2 + 2x)(3 - (x^2)/2) = 6 - x^2 +6x - (x^3)

    = 6 + 6x - x^2 (approx ignoring x^3 and higher)

    sorry ignore that, misread the question, use mclaurin then use binomial expansion and simplify
 
 
 

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