anonymous
  • anonymous
Linear Approximation (.99)^25
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
think of f(x)=x^25 f'(x)=25x^24 linear approximation f(25)= f(1)+ f'(1)(0.99-1)=1-25*0.01=0.75 which is fairly close to the actual value: 0.777821
anonymous
  • anonymous
f(25) that is f(1)
anonymous
  • anonymous
Andras has it. you could also use \[f(x)=(1-x)^{25}\] \[f'(x)=-25(1-x)^{24}\] find the equation of the line tangent to the curve and then let x = .1

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anonymous
  • anonymous
Thanks!!! You guys are awesome!! :D

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