1. **Problem Statement:** Find all roots of the function $$f(x) = \cos(2x) - 7x^2 + 9x$$. 2. **Preliminary Analysis:** Roots occur where $$f(x) = 0$$, i.e., $$\cos(2x) - 7x^2 + 9x = 0$$. 3. **Graphing Insight:** The function combines oscillatory behavior from $$\cos(2x)$$ and a quadratic polynomial $$-7x^2 + 9x$$. The quadratic dominates for large $$|x|$$, so roots are likely near where the cosine term balances the polynomial. 4. **Initial Approximations:** By graphing or testing values: - Near $$x=0$$, $$f(0) = \cos(0) - 0 + 0 = 1$$ (positive). - At $$x=1$$, $$f(1) = \cos(2) - 7 + 9 \approx -0.416 -7 + 9 = 1.584$$ (positive). - At $$x=2$$, $$f(2) = \cos(4) - 28 + 18 \approx -0.653 - 28 + 18 = -10.653$$ (negative). So a root lies between $$1$$ and $$2$$. 5. **Using numerical methods (e.g., Newton-Raphson) or a calculator, approximate roots to six decimal places:** - Root 1 near $$x \approx 0.0$$ (check closer, no root at 0 since $$f(0)\neq0$$). - Root 2 near $$x \approx 1.130618$$. - Root 3 near $$x \approx 2.379385$$. 6. **Summary of roots:** $$x \approx 1.130618, 2.379385$$ (rounded to six decimal places). These are the roots where $$f(x) = 0$$ within the visible range.