anonymous
  • anonymous
I could really use a walk through of the method for solving this Approximate the critical number of f(x) = 18x cos (x) on the interval (0,pi) round to three decimal places.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
first we need the derivative, it is \[18(\cos(x)-x\sin(x))\] by the product rule
anonymous
  • anonymous
oh good I managed to get that far lol but that was pretty much it.
anonymous
  • anonymous
critical point will be the numbers in the interval \((0,\pi)\) where \[\cos(x)-x\sin(x)=0\]

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anonymous
  • anonymous
there is no easy way to solve this
anonymous
  • anonymous
you can try \[\cos(x)=x\sin(x)\] or \[\frac{\cos(x)}{x}=\sin(x)\] but algebra will not get it for you, you need some sort of technology or newton-ralphson method. i would use technology
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=cos%28x%29-xsin%28x%29%3D0
anonymous
  • anonymous
the ugly decimals will show you that there is no snap way to do it
anonymous
  • anonymous
we were working on newton's method I ended up with 1.57 as an approximation of the zero but I was not sure if that was right.
anonymous
  • anonymous
i guess not since that is neither of the answers from wolf
anonymous
  • anonymous
even newton's method will require technolgy because there is no way for you know know what say \[\sin(1)\] is so forget that mess and cheat
anonymous
  • anonymous
lol ok thanks I did get .86 with my calculator initially.
anonymous
  • anonymous
thank you :)
anonymous
  • anonymous
good, others are there as well
anonymous
  • anonymous
yw
anonymous
  • anonymous
the others aren't in my interval.
anonymous
  • anonymous
really interesting...
anonymous
  • anonymous
yeah only one is in the inteval

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