Polynomial Function B1Dd9D

1. **Problem Statement:** We are given three polynomial functions: - $$f(x) = x^4 - x^3 - 4x^2 + 4$$ - $$f(x) = -x^3 + x^2 + 3x - 3$$ - $$y = x^5 - 3x + 1$$ We will choose the first function $$f(x) = x^4 - x^3 - 4x^2 + 4$$ to analyze. 2. **Domain:** The domain of any polynomial function is all real numbers, so: $$\text{Domain} = (-\infty, \infty)$$ 3. **Range:** To find the range, we analyze the behavior of $$f(x)$$. - Since the leading term is $$x^4$$, which grows large and positive as $$x \to \pm \infty$$, the function tends to $$+\infty$$ at both ends. - We find critical points by differentiating: $$f'(x) = 4x^3 - 3x^2 - 8x$$ Set $$f'(x) = 0$$: $$4x^3 - 3x^2 - 8x = 0$$ Factor out $$x$$: $$x(4x^2 - 3x - 8) = 0$$ Solve quadratic $$4x^2 - 3x - 8 = 0$$ using quadratic formula: $$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-8)}}{2 \cdot 4} = \frac{3 \pm \sqrt{9 + 128}}{8} = \frac{3 \pm \sqrt{137}}{8}$$ Approximate roots: $$x \approx 0, \quad x \approx 1.64, \quad x \approx -1.21$$ 4. **Evaluate $$f(x)$$ at critical points:** - $$f(0) = 0 - 0 - 0 + 4 = 4$$ - $$f(1.64) \approx (1.64)^4 - (1.64)^3 - 4(1.64)^2 + 4 \approx 7.23 - 4.41 - 10.75 + 4 = -3.93$$ - $$f(-1.21) \approx (-1.21)^4 - (-1.21)^3 - 4(-1.21)^2 + 4 \approx 2.15 + 1.77 - 5.86 + 4 = 2.06$$ 5. **Range conclusion:** The minimum value is approximately $$-3.93$$ at $$x \approx 1.64$$. Since $$f(x) \to +\infty$$ as $$x \to \pm \infty$$, the range is: $$\text{Range} = [-3.93, \infty)$$ 6. **Create a table of values:** | $$x$$ | $$f(x)$$ | |-------|----------| | -2 | $$16 + 8 - 16 + 4 = 12$$ | | -1 | $$1 + 1 - 4 + 4 = 2$$ | | 0 | $$4$$ | | 1 | $$1 - 1 - 4 + 4 = 0$$ | | 2 | $$16 - 8 - 16 + 4 = -4$$ | 7. **Graph:** The graph is a quartic curve with a local minimum near $$x=1.64$$ at approximately $$-3.93$$, and it rises to infinity on both ends.