The nth term of the Taylor series for f(x) centered at a=1 is given by:
f^(n)(1)/n! * (x-1)^n
where f^(n)(1) is the nth derivative of f evaluated at x=1.
Since f(x) = e^(5x), we have:
f'(x) = 5e^(5x)
f''(x) = 25e^(5x)
f'''(x) = 125e^(5x)
f''''(x) = 625e^(5x)
Evaluating these at x=1, we get:
f'(1) = 5e^5
f''(1) = 25e^5
f'''(1) = 125e^5
f''''(1) = 625e^5
Substituting these into the formula for the 4th Taylor polynomial, we get:
P4(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)^2/2 + f'''(1)(x-1)^3/6 + f''''(1)(x-1)^4/24
= e^5 + 5e^5(x-1) + 25e^5(x-1)^2/2 + 125e^5(x-1)^3/6 + 625e^5(x-1)^4/24
Simplifying, we get:
P4(x) = e^5 * (1 + 5(x-1) + 25(x-1)^2/2 + 125(x-1)^3/6 + 625(x-1)^4/24)