Step 1: Understand the Taylor series expansion.
The Taylor series expansion of a function \( f(x) \) about a point \( x = a \) is: \[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f^{(3)}(a)}{3!}(x - a)^3 + \frac{f^{(4)}(a)}{4!}(x - a)^4 + \cdots, \] \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n, \] where \( f^{(n)}(a) \) is the \( n \)-th derivative of \( f(x) \) evaluated at \( x = a \). Here, \( f(x) = e^x \), and the expansion is about \( x = 2 \), so \( a = 2 \). We need the coefficient of \( (x - 2)^4 \).
Step 2: Compute the derivatives of \( f(x) = e^x \).
\( f(x) = e^x \),
\( f'(x) = e^x \),
\( f''(x) = e^x \),
\( f^{(n)}(x) = e^x \) for all \( n \).
Evaluate at \( x = 2 \): \[ f^{(n)}(2) = e^2. \]
Step 3: Write the Taylor series and find the coefficient.
The Taylor series for \( e^x \) about \( x = 2 \) is: \[ e^x = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!} (x - 2)^n = \sum_{n=0}^{\infty} \frac{e^2}{n!} (x - 2)^n. \] The term with \( (x - 2)^4 \) corresponds to \( n = 4 \). The coefficient of \( (x - 2)^4 \) is: \[ \frac{f^{(4)}(2)}{4!} = \frac{e^2}{4!}. \]
Step 4: Evaluate the options.
(1) \( \frac{1}{4!} \): Incorrect, as the coefficient includes \( e^2 \), not 1. Incorrect.
(2) \( \frac{2^4}{4!} \): Incorrect, as \( 2^4 = 16 \), but the coefficient involves \( e^2 \), not \( 2^4 \). Incorrect.
(3) \( \frac{e^2}{4!} \): Correct, as this matches the coefficient of \( (x - 2)^4 \). Correct.
(4) \( \frac{e^4}{4!} \): Incorrect, as the exponent is \( e^2 \) (since the expansion is about \( x = 2 \)), not \( e^4 \). Incorrect.
Step 5: Select the correct answer.
The coefficient of \( (x - 2)^4 \) is \( \frac{e^2}{4!} \), matching option (3).