1. **Problem:** Find the Maclaurin polynomial for $f(x) = \ln(3 + 4x)$. 2. **Formula and rules:** The Maclaurin polynomial is the Taylor polynomial centered at 0: $$P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k$$ We find derivatives of $f(x)$, evaluate at 0, and form the polynomial. 3. **Step 1: Evaluate $f(0)$:** $$f(0) = \ln(3 + 4\cdot0) = \ln 3$$ 4. **Step 2: First derivative:** $$f'(x) = \frac{4}{3 + 4x}$$ Evaluate at 0: $$f'(0) = \frac{4}{3}$$ 5. **Step 3: Second derivative:** $$f''(x) = -\frac{16}{(3 + 4x)^2}$$ Evaluate at 0: $$f''(0) = -\frac{16}{9}$$ 6. **Step 4: Third derivative:** $$f'''(x) = \frac{128}{(3 + 4x)^3}$$ Evaluate at 0: $$f'''(0) = \frac{128}{27}$$ 7. **Step 5: Write Maclaurin polynomial up to degree 3:** $$P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3$$ Substitute values: $$P_3(x) = \ln 3 + \frac{4}{3}x + \frac{-16/9}{2}x^2 + \frac{128/27}{6}x^3$$ Simplify coefficients: $$P_3(x) = \ln 3 + \frac{4}{3}x - \frac{8}{9}x^2 + \frac{64}{81}x^3$$ **Final answer:** $$\boxed{P_3(x) = \ln 3 + \frac{4}{3}x - \frac{8}{9}x^2 + \frac{64}{81}x^3}$$