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Let [tex]f[/tex] be the function defined by [tex]f(x)=\sqrt{x}[/tex]. What is the approximation for the value of [tex]\sqrt{3}[/tex] obtained by using the second-degree Taylor polynomial for [tex]f[/tex] about [tex]x=4[/tex]?

Respuesta :

Answer:

Step-by-step explanation:

The Taylor series at x=4 is

[tex]\sqrt{x} = 2 + \frac{x-4}{4} - \frac{(x-4)^2}{64} + ...[/tex]

So,

[tex]\sqrt{3} = 2 - \frac{1}{4} - \frac{1}{64} = 1.73[/tex]

Not too bad, eh?

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