annabmovi2clhot6ch annabmovi2clhot6ch
  • 11-11-2016
  • Mathematics
contestada

Estimate sqrt(99.8) using differentials (or equivalently, a linear approximation).

Respuesta :

alsteva alsteva
  • 16-11-2016
The formula for a linearized function about a point x=a is :

[tex]L(x) = f(a)+f'(a)(x-a)[/tex]

Take a = 100

[tex]f(x)= \sqrt{x} \\f'(x)= (\sqrt{x} )'= \frac{1}{2 \sqrt{x} } \\ \\f(100)= \sqrt{100}=10 \\ f'(100)= \frac{1}{2 \sqrt{100} }= \frac{1}{2 \times 10 }= \frac{1}{20 } \\ \\L(99.8)=10 + \frac{1}{20 }(99.8-100)=10- \frac{0.2}{20}= 9.99[/tex] 
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