What is maclaurin 3(1+e^{-0.4t}) ?

The answer to maclaurin 3(1+e^{-0.4t}) is 6-1.2t+0.24t^2-0.032t^3+0.0032t^4+\ldots

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maclaurin 3(1+e^{-0.4t})

maclaurin

3 (1 + e^{- 0.4 t})

6 - 1.2 t + 0.24 t^{2} - 0.032 t^{3} + 0.0032 t^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 6 + \frac{\frac{d}{d t} ( 3 ( 1 + e ^{- 0.4 t} ) ) ( 0 )}{1 !} t + \frac{\frac{d ^{2}}{d t ^{2}} ( 3 ( 1 + e ^{- 0.4 t} ) ) ( 0 )}{2 !} t^{2} + \frac{\frac{d ^{3}}{d t ^{3}} ( 3 ( 1 + e ^{- 0.4 t} ) ) ( 0 )}{3 !} t^{3} + \dots

Evaluate Derivatives

= 6 + \frac{- 1.2}{1 !} t + \frac{0.48}{2 !} t^{2} + \frac{- 0.192}{3 !} t^{3} + \frac{0.0768}{4 !} t^{4} + \dots

Refine

= 6 - 1.2 t + 0.24 t^{2} - 0.032 t^{3} + 0.0032 t^{4} + \dots

y = 5 x - 2 x, (1, 3)

x \to 0 lim (sin (x) - 2 x)

\int 8 x y

y = (1 + 5 x)^{8}, (0, 1)

\frac{d}{d x} (n ln (0.4 n x + 0.6))

What is maclaurin 3(1+e^{-0.4t}) ?
The answer to maclaurin 3(1+e^{-0.4t}) is 6-1.2t+0.24t^2-0.032t^3+0.0032t^4+\ldots