What is taylor e^{-3x+1}+8 ?

The answer to taylor e^{-3x+1}+8 is e+8-3ex+(9e)/2 x^2-(9e)/2 x^3+(27e)/8 x^4+\ldots

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taylor e^{-3x+1}+8

taylor

e^{- 3 x + 1} + 8

e + 8 - 3 e x + \frac{9 e}{2} x^{2} - \frac{9 e}{2} x^{3} + \frac{27 e}{8} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= e + 8 + \frac{\frac{d}{d x} ( e ^{- 3 x + 1} + 8 ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{- 3 x + 1} + 8 ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{- 3 x + 1} + 8 ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= e + 8 + \frac{- 3 e}{1 !} x + \frac{9 e}{2 !} x^{2} + \frac{- 27 e}{3 !} x^{3} + \frac{81 e}{4 !} x^{4} + \dots

Refine

= e + 8 - 3 e x + \frac{9 e}{2} x^{2} - \frac{9 e}{2} x^{3} + \frac{27 e}{8} x^{4} + \dots

\int \frac{e ^{2 t}}{e ^{2 t} - e ^{- 2 t}} d t

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

(sin (x^{2}))^{^{'}}

x - sin (π x)

(x + 8)^{\frac{1}{3}}

What is taylor e^{-3x+1}+8 ?
The answer to taylor e^{-3x+1}+8 is e+8-3ex+(9e)/2 x^2-(9e)/2 x^3+(27e)/8 x^4+\ldots