1. **Problem:** Find all numbers $c$ in $[-1,3]$ such that $f'(c) = \frac{f(3)-f(-1)}{3-(-1)}$ for $f(x) = x^2 - 2x - 8$ (Rolle's Theorem requires $f(-1) = f(3)$). 2. **Check Rolle's Theorem conditions:** - $f$ is continuous and differentiable on $[-1,3]$. - Calculate $f(-1) = (-1)^2 - 2(-1) - 8 = 1 + 2 - 8 = -5$. - Calculate $f(3) = 3^2 - 2(3) - 8 = 9 - 6 - 8 = -5$. - Since $f(-1) = f(3)$, Rolle's Theorem applies. 3. **Find $c$ such that $f'(c) = 0$:** - $f'(x) = 2x - 2$. - Set $2c - 2 = 0 \Rightarrow c = 1$. - Check $c \in [-1,3]$: yes. --- 4. **Problem:** Verify Lagrange's Mean Value Theorem (MVT) for $f(x) = x + \frac{1}{x}$ on $[1,3]$. 5. **Check conditions:** - $f$ is continuous and differentiable on $[1,3]$. 6. **Calculate average rate of change:** - $\frac{f(3) - f(1)}{3 - 1} = \frac{3 + \frac{1}{3} - (1 + 1)}{2} = \frac{3.3333 - 2}{2} = \frac{1.3333}{2} = 0.6667$. 7. **Find $c$ such that $f'(c) = 0.6667$:** - $f'(x) = 1 - \frac{1}{x^2}$. - Set $1 - \frac{1}{c^2} = 0.6667 \Rightarrow \frac{1}{c^2} = 0.3333 \Rightarrow c^2 = 3 \Rightarrow c = \sqrt{3} \approx 1.732$. - $c \in (1,3)$, so MVT holds. --- 8. **Problem:** Find radius of convergence for $\sum_{n=1}^\infty \frac{(x+3)^n}{n^2}$. 9. **Use root test:** - $\lim_{n \to \infty} \sqrt[n]{\left| \frac{(x+3)^n}{n^2} \right|} = \lim_{n \to \infty} \frac{|x+3|}{n^{2/n}} = |x+3|$. 10. **Radius of convergence $R$ satisfies:** - $|x+3| < 1 \Rightarrow R = 1$. --- 11. **Problem:** Find Maclaurin series for $f(x) = \ln(1+x)$, radius and interval of convergence. 12. **Maclaurin series:** - $f(x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}$. 13. **Radius of convergence:** - From geometric series, $R=1$. 14. **Interval of convergence:** - $(-1,1]$. - At $x=-1$, series is $\sum (-1)^{n+1} \frac{(-1)^n}{n} = -\sum \frac{1}{n}$ diverges. - At $x=1$, series is alternating harmonic series which converges. --- 15. **Problem:** For $x>0$, analyze convergence of series $\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1) \sqrt{n}}$ (rewritten from given terms). 16. **Use root or ratio test:** - Dominant term behaves like $\frac{x^{n+1}}{n^{3/2}}$. - Series converges for all $x$ because $p$-series with $p>1$ and exponential term. 17. **More precise:** - Since $x^{n+1}$ grows, convergence depends on $|x|<1$. --- 18. **Problem:** For $\sum_{n=1}^\infty \frac{(x-2)^n}{n 3^n}$, find radius and interval of convergence. 19. **Radius of convergence:** - Use root test: - $\lim_{n \to \infty} \sqrt[n]{\left| \frac{(x-2)^n}{n 3^n} \right|} = \frac{|x-2|}{3}$. - So $R=3$. 20. **Interval:** - $|x-2| < 3 \Rightarrow (-1,5)$. 21. **Check endpoints:** - At $x=-1$, series is $\sum \frac{(-3)^n}{n 3^n} = \sum \frac{(-1)^n}{n}$ converges (alternating harmonic). - At $x=5$, series is $\sum \frac{3^n}{n 3^n} = \sum \frac{1}{n}$ diverges. 22. **Final interval:** $[-1,5)$. **Final answers:** - 1) $c=1$ - 2) $c=\sqrt{3}$ - 3) $R=1$ - 4) Maclaurin series $\sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}$, $R=1$, interval $(-1,1]$ - 5) Converges for $|x|<1$ - 6a) $R=3$ - 6b) Interval $[-1,5)$