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Critical points of Sec x on an interval?

g(x)= sec(x) on the closed interval [ -pi/6, pi/3]..... I got the derivative of Sec x Tan x but I don't know what to do from there

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  • 8 years ago
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       g(x) =  y = sec(x)

     dy ⁄ dx = sec(x) • tan(x) ... set to zero for min/max

          0 = sec(x) • tan(x)

    ... either or both of the terms: sec(x) and tan(x) must be zero.

    First Term Solution:

                 sec(x) = 0

              1 ⁄ cos(x) = 0

     ... no solution since cos(x) never equals infinity

    Second Term Solution:

                  tan(x) = 0

             sin(x) ⁄ cos(x) = 0 ... solution is when numerator is zero

                   sin(x) = 0

                    x = 0 ... on the interval: [- π ⁄ 6, π ⁄ 3 ]

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    2nd derivative:

      d²y ⁄ dx²  =  sec(x) • sec²(x)  +  tan(x) • sec(x) • tan(x)

      d²y ⁄ dx²  =  sec³(x)  +  tan²(x) • sec(x)

      d²y ⁄ dx²  =  sec(x) • [ sec²(x)  +  tan²(x) ]

    at x = 0 :

      d²y ⁄ dx²  =  sec(0) • [ sec²(0)  +  tan²(0) ]

      d²y ⁄ dx²  =  1 • [ 1  +  0 ]

      d²y ⁄ dx²  =  1 ... which is > 0 , so at (x=1) there is a minimum

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    ... at the boundaries of the interval: [- π ⁄ 6, π ⁄ 3 ]

    ... when checking the polarity of the 2nd derivative ONLY,

    ... the bracketed multiplier: [ sec²(- π ⁄ 6)  +  tan²(- π ⁄ 6) ]

    ... can be ignored since it is always > 0

    at x = - π ⁄ 6 : Polarity Check

     d²y ⁄ dx²  =  sec(- π ⁄ 6) • [ sec²(- π ⁄ 6)  +  tan²(- π ⁄ 6) ]

     d²y ⁄ dx²  =  sec(- π ⁄ 6) ... ignoring the bracketed multiplier

         = which is: > 0

    at x = π ⁄ 3 : Polarity Check

     d²y ⁄ dx²  =  sec(π ⁄ 3) • [ sec²(π ⁄ 3)  +  tan²(π ⁄ 3) ]

     d²y ⁄ dx²  =  sec(π ⁄ 3) ... ignoring the bracketed multiplier

         = which is: > 0

    In fact: sec(x) is ALWAYS > 0 , on the interval : [- π ⁄ 6, π ⁄ 3 ]

    ... so the 2nd derivative doesn't change sign ...

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    The only critical point on the given interval is a minimum at x = 0, g(x) = 1

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  • 3 years ago

    enable x be the size and y be the width of the oblong backyard. Perimeter = 2x+2y = 1000 2y=1000-2x y=500-x fee = 2(25)x + 25y + 17y (on condition that fee is $25 for 3 factors and $17 for one ingredient) fee = 50x+42y xy=1000 y=1000/x C = 50 x + 40 two(1000/x) = 50x +40 2,000/x differentiate with take excitement in to x, set it equivalent to 0 and resolve for x dC/dx = 50 - 40 2,000 / x^2 = 0 40 2,000 / x^2 = 50 50x^2 = 40 two,000 x^2 = 840 x=28.ninety 8 y=1000/x = 34.51 d^2C/dx^2 = (-40 2,000)(-2) x^(-3) = eighty 4,000 /x^3 > 0, whilst x=28.ninety 8 This shows the fee has been minimized. minimum fee = 50x+42y = $2,948.40 two

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