1. Stated Problem: Find the function and analyze it as given $$f(x) = 2(x - 3x^2)^4$$. 2. Inside the parentheses, notice that $$x - 3x^2$$ is a quadratic expression. 3. The function is the fourth power of this quadratic, multiplied by 2, which means its output is always non-negative (since any number to the fourth power is non-negative). 4. To analyze, first consider the base function inside: $$g(x) = x - 3x^2$$. 5. This is a downward-opening parabola with vertex at $$x = \frac{-b}{2a} = \frac{-1}{2 \times (-3)} = \frac{1}{6}$$. 6. At $$x = \frac{1}{6}$$, $$g\left(\frac{1}{6}\right) = \frac{1}{6} - 3\left(\frac{1}{6}\right)^2 = \frac{1}{6} - 3 \times \frac{1}{36} = \frac{1}{6} - \frac{1}{12} = \frac{1}{12}$$. 7. Since the original function is $$f(x) = 2(g(x))^4$$, it attains minimum value 0 when $$g(x) = 0\Rightarrow x - 3x^2 = 0 \Rightarrow x(1-3x) = 0 \Rightarrow x = 0, \frac{1}{3}$$. 8. The function has minima at these zeros where $$f(x) = 0$$. 9. Behavior: Since it's a fourth power, the graph touches the x-axis at $$x=0$$ and $$x=\frac{1}{3}$$ but does not cross it. 10. End behavior: As $$x \to \pm \infty$$, $$f(x) \to \infty$$ because the highest degree term dominates. Final answer: The function is $$f(x) = 2(x - 3x^2)^4$$ with zeros at $$x = 0$$ and $$x = \frac{1}{3}$$, minimum values of 0, and grows large positively outside these points.