1. **Problem:** Estimate the roots of the equation $$\cos x - 2x^2 = 0$$ using the Maclaurin polynomial. 2. **Recall the Maclaurin series for $$\cos x$$:** $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$ 3. **Approximate $$\cos x$$ by its second-degree Maclaurin polynomial:** $$P_2(x) = 1 - \frac{x^2}{2}$$ 4. **Substitute into the equation:** $$1 - \frac{x^2}{2} - 2x^2 = 0$$ 5. **Simplify:** $$1 - \frac{x^2}{2} - 2x^2 = 1 - \frac{5x^2}{2} = 0$$ 6. **Solve for $$x^2$$:** $$\frac{5x^2}{2} = 1 \implies x^2 = \frac{2}{5}$$ 7. **Roots:** $$x = \pm \sqrt{\frac{2}{5}} = \pm \frac{\sqrt{10}}{5}$$ --- 1. **Problem:** Estimate the roots of the equation $$x - e^{-x} = 0$$ using the Maclaurin polynomial. 2. **Recall the Maclaurin series for $$e^{-x}$$:** $$e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots$$ 3. **Approximate $$e^{-x}$$ by its second-degree Maclaurin polynomial:** $$P_2(x) = 1 - x + \frac{x^2}{2}$$ 4. **Substitute into the equation:** $$x - \left(1 - x + \frac{x^2}{2}\right) = 0$$ 5. **Simplify:** $$x - 1 + x - \frac{x^2}{2} = 0 \implies 2x - 1 - \frac{x^2}{2} = 0$$ 6. **Multiply both sides by 2:** $$4x - 2 - x^2 = 0 \implies -x^2 + 4x - 2 = 0$$ 7. **Rewrite:** $$x^2 - 4x + 2 = 0$$ 8. **Solve quadratic:** $$x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}$$ --- 1. **Problem:** Estimate the roots of the equation $$\ln(x^2 + 3x + 1) = x$$ using the Maclaurin polynomial. 2. **Let $$f(x) = \ln(x^2 + 3x + 1) - x$$. We want to find roots of $$f(x) = 0$$.** 3. **Expand $$\ln(1 + u)$$ Maclaurin series where $$u = x^2 + 3x$$:** $$\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$$ 4. **Approximate up to second order:** $$\ln(x^2 + 3x + 1) \approx (x^2 + 3x) - \frac{(x^2 + 3x)^2}{2}$$ 5. **Expand:** $$(x^2 + 3x)^2 = x^4 + 6x^3 + 9x^2$$ 6. **Substitute:** $$\ln(x^2 + 3x + 1) \approx x^2 + 3x - \frac{x^4 + 6x^3 + 9x^2}{2}$$ 7. **Simplify:** $$\approx x^2 + 3x - \frac{x^4}{2} - 3x^3 - \frac{9x^2}{2} = 3x + x^2 - \frac{9x^2}{2} - 3x^3 - \frac{x^4}{2}$$ 8. **Combine like terms:** $$3x - \frac{7x^2}{2} - 3x^3 - \frac{x^4}{2}$$ 9. **Set equal to $$x$$:** $$3x - \frac{7x^2}{2} - 3x^3 - \frac{x^4}{2} = x$$ 10. **Bring all terms to one side:** $$3x - x - \frac{7x^2}{2} - 3x^3 - \frac{x^4}{2} = 0 \implies 2x - \frac{7x^2}{2} - 3x^3 - \frac{x^4}{2} = 0$$ 11. **Multiply entire equation by 2:** $$4x - 7x^2 - 6x^3 - x^4 = 0$$ 12. **Factor out $$x$$:** $$x(4 - 7x - 6x^2 - x^3) = 0$$ 13. **One root is $$x=0$$. For others, solve:** $$4 - 7x - 6x^2 - x^3 = 0$$ 14. **This cubic can be approximated numerically or by trial. For small $$x$$, the root near zero is $$x=0$$.** --- 1. **Problem:** Estimate the roots of the equation $$e^{\sqrt{x}} - 3x = -5$$ using the Maclaurin polynomial. 2. **Rewrite:** $$e^{\sqrt{x}} = 3x - 5$$ 3. **Let $$t = \sqrt{x}$$, then $$x = t^2$$, so:** $$e^t = 3t^2 - 5$$ 4. **Maclaurin series for $$e^t$$:** $$e^t = 1 + t + \frac{t^2}{2} + \frac{t^3}{6} + \cdots$$ 5. **Approximate up to second degree:** $$e^t \approx 1 + t + \frac{t^2}{2}$$ 6. **Substitute:** $$1 + t + \frac{t^2}{2} = 3t^2 - 5$$ 7. **Bring all terms to one side:** $$1 + t + \frac{t^2}{2} - 3t^2 + 5 = 0 \implies 6 + t - \frac{5t^2}{2} = 0$$ 8. **Multiply by 2:** $$12 + 2t - 5t^2 = 0$$ 9. **Rewrite:** $$-5t^2 + 2t + 12 = 0$$ 10. **Multiply by -1:** $$5t^2 - 2t - 12 = 0$$ 11. **Solve quadratic:** $$t = \frac{2 \pm \sqrt{4 + 240}}{10} = \frac{2 \pm \sqrt{244}}{10} = \frac{2 \pm 2\sqrt{61}}{10} = \frac{1 \pm \sqrt{61}}{5}$$ 12. **Since $$t = \sqrt{x} \geq 0$$, take positive root:** $$t = \frac{1 + \sqrt{61}}{5}$$ 13. **Find $$x$$:** $$x = t^2 = \left(\frac{1 + \sqrt{61}}{5}\right)^2$$ --- **Final answers:** 1. $$x = \pm \frac{\sqrt{10}}{5}$$ 2. $$x = 2 \pm \sqrt{2}$$ 3. $$x = 0$$ (and other roots from cubic approximation) 4. $$x = \left(\frac{1 + \sqrt{61}}{5}\right)^2$$