taylor x^{1/3},x=8

Lösung

taylor

x^{\frac{1}{3}}, x = 8

2 + \frac{1}{12} (x - 8) - \frac{1}{288} (x - 8)^{2} + \frac{5}{20736} (x - 8)^{3} - \frac{5}{248832} (x - 8)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 2 + \frac{\frac{1}{12}}{1 !} (x - 8) + \frac{- \frac{1}{144}}{2 !} (x - 8)^{2} + \frac{\frac{5}{3456}}{3 !} (x - 8)^{3} + \frac{- \frac{5}{10368}}{4 !} (x - 8)^{4} + \dots

Fasse zusammen

= 2 + \frac{1}{12} (x - 8) - \frac{1}{288} (x - 8)^{2} + \frac{5}{20736} (x - 8)^{3} - \frac{5}{248832} (x - 8)^{4} + \dots

x \to 0 lim (\frac{- x ^{3}}{x ^{2}})

t an g e n t f (x) = a x^{2}, at x = - 2

\int \frac{2 x ^{3} + 3 x ^{2} + x + 1}{2 x + 1} d x

a re a y = 2 x + 5, y = 15 - x^{2}, x = - 1, x = 2

\int (4 + 3 x)^{5} d x