anonymous
  • anonymous
f(x) = e^(2x) + e^(−2x) Find an upper bound for the error in using the second degree Maclaurin polynomial of f to approximate f(0.5).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I guess I have to use the Lagrange remainder formula, right? \[E _{n}(x) = f(x) - T _{n}(x) =\frac{ f ^{(n+1)}(c) }{ (n+1)! }(x-a)^{(n+1)}\] Where c lies between x and a
anonymous
  • anonymous
yes just plug the rest in
anonymous
  • anonymous
So here a=0 right?

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anonymous
  • anonymous
i think so
anonymous
  • anonymous
Is my 3rd derivative correct? I got... \[8e ^{2x}-8e ^{-2x}\]
anonymous
  • anonymous
ya
anonymous
  • anonymous
Cool, so then I get \[E _{2}(0.5)=\frac{ e ^{2c} -e ^{-2c}}{ 6 }\]
anonymous
  • anonymous
yes! im surprised u asked for help cuz u get it
anonymous
  • anonymous
Lol I don't know what to do next :D
anonymous
  • anonymous
that is f(x)
anonymous
  • anonymous
i see what u do now
anonymous
  • anonymous
u have to
anonymous
  • anonymous
replace x with .5
anonymous
  • anonymous
What if I was trying to find the lower bound?

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