Find the Maclaurin polynomial of degree 5 for the following function: F(x)=e^(-x)?

Using the series e^t = sum(n=0 to infinity) t^n/n!,
replace t with (-x):

e^(-x) = sum(n=0 to infinity) (-t)^k/k!
= 1 - t + t^2/2 - t^3/3! + t^4/4! - t^5/5! + ...
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Or, directly from the definition:
f(x) = e^(-x) ==> f(0) = 1
f'(x) = -e^(-x) ==> f'(0) = -1
f''(x) = e^(-x) ==> f''(0) = 1
f'''(x) = -e^(-x) ==> f'''(0) = -1
f''''(x) = e^(-x) ==> f''''(0) = 1
f'''''(x) = -e^(-x) ==> f'''''(0) = -1.

So,
f(x) = e^(-x)
= sum(n=0 to infinity) f^(k)(0) (x - 0)^k/k!
= 1 - t + t^2/2 - t^3/3! + t^4/4! - t^5/5! + ...

I hope this helps!

f ( x ) = e^( - 4x) = a million f ' ( x ) = - 4e^( - 4x ) = - 4 f '' ( x ) = 16e^( - 4x ) = sixteen f ''' ( x ) = - 64e^( - 4x ) = - sixty 4 f^( iv ) ( x ) = 256e^( - 4x ) = 256 f^(v) ( x ) = - 1024e^( - 4x ) = - 1024 (a million/ 0!) - (4x / a million!) + (16x^2 / 2!) - (64x^3 / 3! )+ (256x^4 / 4!) - (1024x^5 / 5!)

e^(0) = 1 so

F(x) = 1 - x + x^2/2 - x^3/6 + x^4/24 - x^5/120