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SquareRoot (Taylor Series)

Calculus Level 2

Determine the first 3 terms of the Taylor series for the function \[f(x) = \sqrt{x} \text{ centered at } x = 9.\]

\[3 + \frac{1}{6}(x - 9) - \frac{1}{108}(x - 9)^2\] \[3 + \frac{1}{6}(x - 9) + \frac{1}{108}(x - 9)^2\] \[3 + \frac{1}{6}(x - 9) - \frac{1}{216}(x - 9)^2\] \[3 + \frac{1}{6}(x - 9) + \frac{1}{216}(x - 9)^2\]

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by Andrew Ellinor

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