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Chapter 9: Problem 27

Finding a Taylor Polynomial In Exercises \(25-30\) , find the \(n\)th Taylor polynomial centered at \(c .\) $$ f(x)=\sqrt{x}, \quad n=3, \quad c=4 $$

Short Answer

Expert verified
The 3rd degree Taylor polynomial centred at \(c = 4\) for the function \(f(x) = \sqrt{x}\) is \(P_3(x) = 2 + \frac{1}{4}(x-4) - \frac{1}{32}(x-4)^2 + \frac{1}{64}(x-4)^3.\

Step by step solution

01

Find the Function Value at c

The function is \(f(x) = \sqrt{x}.\) Thus, \(F(c) = f(4) = \sqrt{4} = 2.\)
02

Calculate the First Derivative and its Value at c

The first derivative of the function is obtained using the power rule: \(f'(x) = \frac{1}{2\sqrt{x}}.\) Thus, \(f'(c) = f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}.\)
03

Calculate the Second Derivative and its Value at c

Next, calculate the second derivative: \(f''(x) = -\frac{1}{4\sqrt{x^3}}.\) Thus, \(f''(c) = f''(4) = -\frac{1}{4\sqrt{4^3}} = -\frac{1}{32}.\)
04

Calculate the Third Derivative and its Value at c

Next, calculate the third derivative: \(f'''(x) = \frac{3}{8\sqrt{x^5}}.\) Thus, \(f'''(c) = f'''(4) = \frac{3}{8\sqrt{4^5}} = \frac{3}{256}.\)
05

Form the Taylor Polynomial

Using the Taylor series formula for \(n=3\), the polynomial is \(P_3(x) = f(4) + f'(4)(x-4) + \frac{f''(4)}{2!}(x-4)^2 + \frac{f'''(4)}{3!}(x-4)^3 = 2 + \frac{1}{4}(x-4) - \frac{1}{32}(x-4)^2 + \frac{3}{256}*(3)(x-4)^3.\

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