What is maclaurin-x^4+2x^3-5x^2+0.4 ?

The answer to maclaurin-x^4+2x^3-5x^2+0.4 is 0.4-5x^2+2x^3-x^4

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maclaurin-x^4+2x^3-5x^2+0.4

maclaurin

- x^{4} + 2 x^{3} - 5 x^{2} + 0.4

0.4 - 5 x^{2} + 2 x^{3} - x^{4}

Solution steps

Apply the Maclaurin Formula

= 0.4 + \frac{\frac{d}{d x} ( - x ^{4} + 2 x ^{3} - 5 x ^{2} + 0.4 ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( - x ^{4} + 2 x ^{3} - 5 x ^{2} + 0.4 ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( - x ^{4} + 2 x ^{3} - 5 x ^{2} + 0.4 ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0.4 + \frac{0}{1 !} x + \frac{- 10}{2 !} x^{2} + \frac{12}{3 !} x^{3} + \frac{- 24}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{0}{10 !} x^{10}

Refine

= 0.4 - 5 x^{2} + 2 x^{3} - x^{4}

\int_{1}^{4} 3 x d x

\frac{s + 5}{s ( s + 10 ) ^{2}}

\frac{x y ^{^{'}}}{e ^{y} + 1} = 2

\int_{- 3}^{1} \frac{x}{9 - x ^{2}} d x

\int \frac{e ^{2 x}}{e ^{2 x} + 6} d x

What is maclaurin-x^4+2x^3-5x^2+0.4 ?
The answer to maclaurin-x^4+2x^3-5x^2+0.4 is 0.4-5x^2+2x^3-x^4