Final answer:
The Taylor polynomial T3(x) simplifies to e + e(x-1) + (e/2!)(x-1)^2 + (e/3!)(x-1)^3.
Step-by-step explanation:
To find the Taylor polynomial T3(x) for the function f(x) = e^x centered at a = 1, we need to use the formula for the general term of the Taylor series expansion:
Tn(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + ... + (f^n(a)/n!)(x-a)^n
Using a = 1 and n = 3, we have:
T3(x) = f(1) + f'(1)(x-1) + (f''(1)/2!)(x-1)^2 + (f^3(1)/3!)(x-1)^3
The function e^x has a constant value of e at every point.
Taking derivatives of e^x repeatedly, we find that the derivatives evaluated at x = 1 are also equal to e.
Therefore, the Taylor polynomial T3(x) for f(x) = e^x centered at a = 1 simplifies to:
T3(x) = e + e(x-1) + (e/2!)(x-1)^2 + (e/3!)(x-1)^3