How do you find the first terms of the taylor series #f(x)=1/sqrt(x)# centered at #x=4#?
From there on, it's just arithmetic.
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To find the first terms of the Taylor series for ( f(x) = \frac{1}{\sqrt{x}} ) centered at ( x = 4 ), we first find the derivatives of ( f(x) ) and evaluate them at ( x = 4 ). Then, we express the Taylor series using these derivatives.

Find the derivatives of ( f(x) ): [ f(x) = x^{\frac{1}{2}} ] [ f'(x) = \frac{1}{2}x^{\frac{3}{2}} ] [ f''(x) = \frac{3}{4}x^{\frac{5}{2}} ] [ f'''(x) = \frac{15}{8}x^{\frac{7}{2}} ]

Evaluate the derivatives at ( x = 4 ): [ f(4) = \frac{1}{2} ] [ f'(4) = \frac{1}{16} ] [ f''(4) = \frac{3}{128} ] [ f'''(4) = \frac{15}{1024} ]

Express the Taylor series: [ f(x) = f(4) + f'(4)(x4) + \frac{f''(4)}{2!}(x4)^2 + \frac{f'''(4)}{3!}(x4)^3 + \ldots ]
Substitute the evaluated derivatives: [ f(x) = \frac{1}{2}  \frac{1}{16}(x4) + \frac{3}{128 \cdot 2!}(x4)^2  \frac{15}{1024 \cdot 3!}(x4)^3 + \ldots ]
These are the first terms of the Taylor series for ( f(x) = \frac{1}{\sqrt{x}} ) centered at ( x = 4 ).
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To find the first terms of the Taylor series ( f(x) = \frac{1}{\sqrt{x}} ) centered at ( x = 4 ), we first need to find the derivatives of ( f(x) ) at ( x = 4 ). Then, we evaluate these derivatives at ( x = 4 ) to find the coefficients of the Taylor series.

Find the derivatives of ( f(x) ): [ f(x) = x^{\frac{1}{2}} ] [ f'(x) = \frac{1}{2}x^{\frac{3}{2}} ] [ f''(x) = \frac{3}{4}x^{\frac{5}{2}} ] [ f'''(x) = \frac{15}{8}x^{\frac{7}{2}} ]

Evaluate the derivatives at ( x = 4 ): [ f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} ] [ f'(4) = \frac{1}{2}(4)^{\frac{3}{2}} = \frac{1}{4} ] [ f''(4) = \frac{3}{4}(4)^{\frac{5}{2}} = \frac{3}{32} ] [ f'''(4) = \frac{15}{8}(4)^{\frac{7}{2}} = \frac{15}{128} ]

Write down the Taylor series using these derivatives: [ f(x) = f(a) + f'(a)(xa) + \frac{f''(a)}{2!}(xa)^2 + \frac{f'''(a)}{3!}(xa)^3 + \cdots ]
[ f(x) = \frac{1}{2}  \frac{1}{4}(x4) + \frac{3}{32}(x4)^2  \frac{15}{128}(x4)^3 + \cdots ]
Therefore, the first terms of the Taylor series for ( f(x) = \frac{1}{\sqrt{x}} ) centered at ( x = 4 ) are:
[ \frac{1}{2}  \frac{1}{4}(x4) + \frac{3}{32}(x4)^2  \frac{15}{128}(x4)^3 ]
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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