1. **Problem statement:** Given the function $f(x) = \frac{7}{2 + 4x^2}$, find its power series representation, interval of convergence, series for its derivative, series for its definite integral from 0 to $\frac{1}{2}$, and determine how many terms are needed to estimate the integral with error less than 0.05. 2. **Rewrite the function:** $$f(x) = \frac{7}{2 + 4x^2} = \frac{7}{2(1 + 2x^2)} = \frac{7}{2} \cdot \frac{1}{1 + 2x^2}$$ 3. **Power series formula:** For $|r| < 1$, $$\frac{1}{1 - r} = \sum_{n=0}^\infty r^n$$ We use this by rewriting denominator as $1 - (-2x^2)$: $$\frac{1}{1 + 2x^2} = \frac{1}{1 - (-2x^2)} = \sum_{n=0}^\infty (-2x^2)^n = \sum_{n=0}^\infty (-1)^n 2^n x^{2n}$$ 4. **Power series for $f(x)$:** $$f(x) = \frac{7}{2} \sum_{n=0}^\infty (-1)^n 2^n x^{2n} = \sum_{n=0}^\infty \frac{7}{2} (-1)^n 2^n x^{2n} = \sum_{n=0}^\infty 7 (-1)^n 2^{n-1} x^{2n}$$ 5. **Interval of convergence:** The series converges when $$| -2x^2 | < 1 \implies 2x^2 < 1 \implies |x| < \frac{1}{\sqrt{2}}$$ 6. **Series for $f'(x)$:** Differentiate term-by-term: $$f'(x) = \frac{d}{dx} \sum_{n=0}^\infty 7 (-1)^n 2^{n-1} x^{2n} = \sum_{n=0}^\infty 7 (-1)^n 2^{n-1} \cdot 2n x^{2n-1} = \sum_{n=1}^\infty 7 (-1)^n 2^{n-1} 2n x^{2n-1}$$ Simplify coefficients: $$f'(x) = \sum_{n=1}^\infty 7 (-1)^n n 2^n x^{2n-1}$$ 7. **Series for $\int_0^{1/2} f(x) dx$:** Integrate term-by-term: $$\int_0^{1/2} f(x) dx = \int_0^{1/2} \sum_{n=0}^\infty 7 (-1)^n 2^{n-1} x^{2n} dx = \sum_{n=0}^\infty 7 (-1)^n 2^{n-1} \int_0^{1/2} x^{2n} dx$$ Integral: $$\int_0^{1/2} x^{2n} dx = \frac{(1/2)^{2n+1}}{2n+1}$$ So, $$\int_0^{1/2} f(x) dx = \sum_{n=0}^\infty 7 (-1)^n 2^{n-1} \frac{(1/2)^{2n+1}}{2n+1} = \sum_{n=0}^\infty 7 (-1)^n 2^{n-1} \frac{1}{2^{2n+1}} \frac{1}{2n+1}$$ Simplify powers of 2: $$2^{n-1} \cdot \frac{1}{2^{2n+1}} = 2^{n-1-(2n+1)} = 2^{-n-2} = \frac{1}{2^{n+2}}$$ Hence, $$\int_0^{1/2} f(x) dx = \sum_{n=0}^\infty \frac{7 (-1)^n}{2^{n+2} (2n+1)}$$ 8. **Number of terms for error < 0.05:** The series is alternating with decreasing terms in absolute value. The error bound is less than the first omitted term. We want smallest $N$ such that $$\left| \frac{7}{2^{N+2} (2N+1)} \right| < 0.05$$ Try $N=2$: $$\frac{7}{2^{4} \cdot 5} = \frac{7}{16 \cdot 5} = \frac{7}{80} = 0.0875 > 0.05$$ Try $N=3$: $$\frac{7}{2^{5} \cdot 7} = \frac{7}{32 \cdot 7} = \frac{7}{224} \approx 0.03125 < 0.05$$ So at least 4 terms (up to $n=3$) are needed. 9. **Practicality:** Calculating 4 terms by hand is practical, but more terms would be tedious. **Final answers:** (a) $$f(x) = \sum_{n=0}^\infty 7 (-1)^n 2^{n-1} x^{2n}$$ (b) Interval of convergence: $$|x| < \frac{1}{\sqrt{2}}$$ (c) $$f'(x) = \sum_{n=1}^\infty 7 (-1)^n n 2^n x^{2n-1}$$ (d) $$\int_0^{1/2} f(x) dx = \sum_{n=0}^\infty \frac{7 (-1)^n}{2^{n+2} (2n+1)}$$ (e) At least 4 terms needed for error less than 0.05; practical to estimate by hand for this accuracy.