What is taylor 3+4x^2-5x^3 ?

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

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taylor 3+4x^2-5x^3

taylor

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

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 3 + \frac{0}{1 !} x + \frac{8}{2 !} x^{2} + \frac{- 30}{3 !} x^{3} + \frac{0}{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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 e^{- 3 x}

\int \frac{1}{1 + \frac{x}{2}} d x

n = 1 \sum \infty \frac{3 ^{n}}{4 ^{n + 1}}

x \to 1 lim (4 x - 1)

\frac{d}{d x} (\frac{( - ln ( 1 + 6 x ) )}{x})