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Math Help - Find points of inflection of trig function

  1. #1
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    Find points of inflection of trig function

    y=sinx-tanx

    Here's my attempt at the derivatives

    y ' = cosx-(sec(x))^2
    y '' = -sinx-2(secx)*secxtanx

    How can I find the points of inflection from that second derivative (where it is zero?)?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Talking Ok

    Quote Originally Posted by theowne View Post
    y=sinx-tanx

    Here's my attempt at the derivatives

    y ' = cosx-(sec(x))^2
    y '' = -sinx-2(secx)*secxtanx

    How can I find the points of inflection from that second derivative (where it is zero?)?
    Check your work against mine... y=sin(x)-tan(x)...so y'=cos(x)-sec(x)^2 and [/tex]-sin(x)-2sec(x)^2tan(x)[/tex]....you cant find the inflection points without an interval but I will assume it is [0,2\pi]...what I would do is graph the curve and find the points of intersection or do this factor out a sec(x)^2 to get -sec(x)^2[sin(x)^3+2tan(x)]=0 then go -sec(x)^2[sin(x)[sin(x)^2+csc(x)]] and work with that
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Oh

    And I almost forgot...an inflection point occurs where the second derivative is zero or undefined...but set up test intervals based on where the second derivative is undefined, zero, and where the original function is undefined...now test an element of each in the second derivative...where there is a sign change between intervals there is an inflection point
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