anonymous
  • anonymous
If the local linear approximation of f(x) = 2cos x + e2x at x = 2 is used to find the approximation for f(2.1), then the % error of this approximation is greater than 15% between 11% and 15% between 5% and 10% between 0% and 4
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
how far did you get here?
anonymous
  • anonymous
well i found the linear approximation for f(2.1) using x = 2

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jim_thompson5910
  • jim_thompson5910
what did you get for f ' (x)
anonymous
  • anonymous
tell me if i did thi right f(2.1) = 53.76 + 107.38(x - 2) f(2.1) = 107.38x - 161
jim_thompson5910
  • jim_thompson5910
I'm getting y = 107.37771x - 160.98955 for the tangent line
jim_thompson5910
  • jim_thompson5910
let g(x) = 107.37771x - 160.98955 compute g(2.1) and compare that with f(2.1)
anonymous
  • anonymous
g(2.1) = 64.503641 f(2.1) = 65.67663883
jim_thompson5910
  • jim_thompson5910
now compute the percent error using those f(2.1) and g(2.1) values
anonymous
  • anonymous
so just divide them by each other?
jim_thompson5910
  • jim_thompson5910
\[\Large \text{Percent Error} = \frac{|\text{Approximate Value}-\text{Exact Value}|}{\text{Exact Value}}\] \[\Large \text{Percent Error} = \frac{|g(2.1)-f(2.1)|}{f(2.1)}\] \[\Large \text{Percent Error} = ??\]
anonymous
  • anonymous
1.81
anonymous
  • anonymous
are you sure it's not backwards on the numerator
anonymous
  • anonymous
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jim_thompson5910
  • jim_thompson5910
the order in the numerator does not matter because |x - y| = |y - x|
jim_thompson5910
  • jim_thompson5910
if the absolute value bars weren't there, then yeah, the order matters
anonymous
  • anonymous
okay well thank you very much :D!!!
jim_thompson5910
  • jim_thompson5910
I'm getting roughly 0.01786 which is approximately 1.786% error
anonymous
  • anonymous
yeah me too either way its D
jim_thompson5910
  • jim_thompson5910
yes

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