What is maclaurin 1/(x-2) ?

The answer to maclaurin 1/(x-2) is -1/2-1/4 x-1/8 x^2-1/16 x^3-1/32 x^4+\ldots

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maclaurin 1/(x-2)

maclaurin

\frac{1}{x - 2}

- \frac{1}{2} - \frac{1}{4} x - \frac{1}{8} x^{2} - \frac{1}{16} x^{3} - \frac{1}{32} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= - \frac{1}{2} + \frac{\frac{d}{d x} ( \frac{1}{x - 2} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x - 2} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x - 2} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= - \frac{1}{2} + \frac{- \frac{1}{4}}{1 !} x + \frac{- \frac{1}{4}}{2 !} x^{2} + \frac{- \frac{3}{8}}{3 !} x^{3} + \frac{- \frac{3}{4}}{4 !} x^{4} + \dots

Refine

= - \frac{1}{2} - \frac{1}{4} x - \frac{1}{8} x^{2} - \frac{1}{16} x^{3} - \frac{1}{32} x^{4} + \dots

f (x) = e^{4 x}

\frac{d ^{2}}{d x ^{2}} (x^{\frac{8}{9}} + 5 x)

\int_{3}^{\infty} \frac{1}{x} d x

\int_{0}^{4} (x^{2} - 4 x) d x

y = ln (- 7 x^{6})

What is maclaurin 1/(x-2) ?
The answer to maclaurin 1/(x-2) is -1/2-1/4 x-1/8 x^2-1/16 x^3-1/32 x^4+\ldots