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

Lösung

taylor

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

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 5 + \frac{4}{1 !} X + \frac{- 2}{2 !} X^{2} + \frac{6}{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}

Fasse zusammen

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

t an g e n t f (x) = \frac{125}{x ^{2} + 24}, at x = - 1

\int \frac{1}{2 x ^{2} - 4 x + 20} d x

\frac{\partial}{\partial x} (x^{2} e^{y z^{2}})

x \to - 3.1 lim (x^{2})

\frac{d y}{d x} = \frac{x x ^{2} - y ^{2} + y ^{2}}{x y}