1. **State the problem:** We want to expand the function $$f(x) = (1+x)^{\frac{5}{2}}$$ using Maclaurin series up to the term in $$x^4$$ and then approximate $$f(0.2)$$. 2. **Recall Maclaurin series formula:** The Maclaurin series expansion of a function $$f(x)$$ is given by $$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \cdots $$ 3. **Calculate derivatives:** - $$f(x) = (1+x)^{\frac{5}{2}}$$ - $$f(0) = 1^{\frac{5}{2}} = 1$$ First derivative: $$ f'(x) = \frac{5}{2}(1+x)^{\frac{3}{2}}$$ $$ f'(0) = \frac{5}{2} \times 1^{\frac{3}{2}} = \frac{5}{2}$$ Second derivative: $$ f''(x) = \frac{5}{2} \times \frac{3}{2} (1+x)^{\frac{1}{2}} = \frac{15}{4}(1+x)^{\frac{1}{2}}$$ $$ f''(0) = \frac{15}{4} \times 1 = \frac{15}{4}$$ Third derivative: $$ f^{(3)}(x) = \frac{15}{4} \times \frac{1}{2} (1+x)^{-\frac{1}{2}} = \frac{15}{8}(1+x)^{-\frac{1}{2}}$$ $$ f^{(3)}(0) = \frac{15}{8} \times 1 = \frac{15}{8}$$ Fourth derivative: $$ f^{(4)}(x) = \frac{15}{8} \times \left(-\frac{1}{2}\right)(1+x)^{-\frac{3}{2}} = -\frac{15}{16}(1+x)^{-\frac{3}{2}}$$ $$ f^{(4)}(0) = -\frac{15}{16} \times 1 = -\frac{15}{16}$$ 4. **Write the Maclaurin expansion up to $$x^4$$:** $$ f(x) \approx 1 + \frac{5}{2}x + \frac{\frac{15}{4}}{2!}x^2 + \frac{\frac{15}{8}}{3!}x^3 + \frac{-\frac{15}{16}}{4!}x^4 $$ 5. **Simplify coefficients:** - $$\frac{15}{4} \div 2! = \frac{15}{4} \times \frac{1}{2} = \frac{15}{8}$$ - $$\frac{15}{8} \div 3! = \frac{15}{8} \times \frac{1}{6} = \frac{15}{48} = \frac{5}{16}$$ - $$-\frac{15}{16} \div 4! = -\frac{15}{16} \times \frac{1}{24} = -\frac{15}{384} = -\frac{5}{128}$$ So, $$ f(x) \approx 1 + \frac{5}{2}x + \frac{15}{8}x^2 + \frac{5}{16}x^3 - \frac{5}{128}x^4 $$ 6. **Calculate approximate value at $$x=0.2$$:** $$ f(0.2) \approx 1 + \frac{5}{2} \times 0.2 + \frac{15}{8} \times 0.2^2 + \frac{5}{16} \times 0.2^3 - \frac{5}{128} \times 0.2^4 $$ Calculate each term: - $$\frac{5}{2} \times 0.2 = 0.5$$ - $$\frac{15}{8} \times 0.04 = \frac{15}{8} \times 0.04 = 0.075$$ - $$\frac{5}{16} \times 0.008 = 0.0025$$ - $$-\frac{5}{128} \times 0.0016 = -0.0000625$$ Sum all: $$ 1 + 0.5 + 0.075 + 0.0025 - 0.0000625 = 1.5774375 $$ Rounded to 4 decimal places: $$ 1.5774 $$