What is taylor x^3+8x-1,-1 ?

The answer to taylor x^3+8x-1,-1 is -10+11(x+1)-3(x+1)^2+(x+1)^3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor x^3+8x-1,-1

taylor

x^{3} + 8 x - 1, - 1

- 10 + 11 (x + 1) - 3 (x + 1)^{2} + (x + 1)^{3}

Solution steps

Apply the Taylor Formula

= - 10 + \frac{\frac{d}{d x} ( x ^{3} + 8 x - 1 ) ( - 1 )}{1 !} (x + 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{3} + 8 x - 1 ) ( - 1 )}{2 !} (x + 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{3} + 8 x - 1 ) ( - 1 )}{3 !} (x + 1)^{3} + \dots

Evaluate Derivatives

= - 10 + \frac{11}{1 !} (x + 1) + \frac{- 6}{2 !} (x + 1)^{2} + \frac{6}{3 !} (x + 1)^{3} + \frac{0}{4 !} (x + 1)^{4} + \frac{0}{5 !} (x + 1)^{5} + \frac{0}{6 !} (x + 1)^{6} + \frac{0}{7 !} (x + 1)^{7} + \frac{0}{8 !} (x + 1)^{8} + \frac{0}{9 !} (x + 1)^{9} + \frac{0}{10 !} (x + 1)^{10}

Refine

= - 10 + 11 (x + 1) - 3 (x + 1)^{2} + (x + 1)^{3}

\frac{\partial}{\partial y} (29 - x^{2} - 4 y^{2})

\frac{d}{d x} (4^{x} lo g_{4} (x))

F (x) = e^{x^{2}}

\frac{d}{d x} (- x^{2})

\int_{0}^{\frac{π}{6}} cos (8 x) d x

What is taylor x^3+8x-1,-1 ?
The answer to taylor x^3+8x-1,-1 is -10+11(x+1)-3(x+1)^2+(x+1)^3