3. Question 3 [6 marks] Let \( f \) be a function with McLaurin series converging to \( f(x) \) for all real numbers \( x \). \( \operatorname{If} f(0)=f^{\prime}(0)=2 \) and \( f^{(n)}(0)=3 \) for \( n \geq 2 \), show \( f(x)=3 e^{x}-x-1 \).

To determine the function \( f(x) \) given its Maclaurin series and the specified derivative values, let's follow these steps: ### 1. **Understand the Maclaurin Series** The Maclaurin series of a function \( f(x) \) is its Taylor series expansion around \( x = 0 \): \[ f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] ### 2. **Plug in the Given Derivative Values** You're provided with the following information: - \( f(0) = 2 \) - \( f'(0) = 2 \) - For all \( n \geq 2 \), \( f^{(n)}(0) = 3 \) Using these, write the Maclaurin series for \( f(x) \): \[ f(x) = f(0) + f'(0) x + \sum_{n=2}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] Substituting the given values: \[ f(x) = 2 + 2x + \sum_{n=2}^{\infty} \frac{3}{n!} x^n \] ### 3. **Simplify the Series** Factor out the constant \( 3 \) from the summation: \[ f(x) = 2 + 2x + 3 \sum_{n=2}^{\infty} \frac{x^n}{n!} \] Notice that the sum \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) is the Maclaurin series for \( e^x \). Therefore, we can express the summation from \( n=2 \) as: \[ \sum_{n=2}^{\infty} \frac{x^n}{n!} = e^x - \frac{x^0}{0!} - \frac{x^1}{1!} = e^x - 1 - x \] Substitute this back into the expression for \( f(x) \): \[ f(x) = 2 + 2x + 3(e^x - 1 - x) \] ### 4. **Combine Like Terms** Distribute the \( 3 \) and combine like terms: \[ f(x) = 2 + 2x + 3e^x - 3 - 3x = 3e^x - x - 1 \] ### 5. **Final Result** Thus, the function \( f(x) \) is: \[ f(x) = 3e^x - x - 1 \]