1. **State the problem:** Analyze and sketch the graph of $$f(x) = x^4 - 3x^2 + 2$$ including domain, range, asymptotes, zeros, multiplicity, factored form, intercepts, derivatives, critical points, concavity, symmetry, and apply Rolle's theorem on $$[-2,2]$$. 2. **Domain:** Polynomial functions are defined for all real numbers, so $$\text{Domain} = (-\infty, \infty)$$. 3. **Factored form and zeros:** $$f(x) = x^4 - 3x^2 + 2 = (x^2)^2 - 3x^2 + 2$$ Let $$u = x^2$$, then $$u^2 - 3u + 2 = 0$$ Factor: $$ (u - 1)(u - 2) = 0$$ So $$u = 1$$ or $$u = 2$$ Back to $$x$$: $$x^2 = 1 \Rightarrow x = \pm 1$$ $$x^2 = 2 \Rightarrow x = \pm \sqrt{2}$$ 4. **Multiplicity:** Each root comes from a linear factor in $$u$$, so each root has multiplicity 1. 5. **Factored form in terms of $$x$$:** $$f(x) = (x^2 - 1)(x^2 - 2) = (x - 1)(x + 1)(x - \sqrt{2})(x + \sqrt{2})$$ 6. **Y-intercept:** Evaluate $$f(0)$$: $$f(0) = 0 - 0 + 2 = 2$$ 7. **Vertical asymptotes:** None, since $$f(x)$$ is a polynomial. 8. **Range:** Analyze critical points to find min/max values. 9. **First derivative:** $$f'(x) = 4x^3 - 6x$$ 10. **Critical numbers:** Solve $$f'(x) = 0$$: $$4x^3 - 6x = 0$$ $$2x(2x^2 - 3) = 0$$ So $$x = 0$$ or $$2x^2 - 3 = 0 \Rightarrow x^2 = \frac{3}{2} \Rightarrow x = \pm \sqrt{\frac{3}{2}}$$ 11. **Increasing/decreasing test:** - For $$x < -\sqrt{\frac{3}{2}}$$, pick $$x = -2$$: $$f'(-2) = 4(-8) - 6(-2) = -32 + 12 = -20 < 0$$ (decreasing) - Between $$-\sqrt{\frac{3}{2}}$$ and 0, pick $$x = -1$$: $$f'(-1) = 4(-1) - 6(-1) = -4 + 6 = 2 > 0$$ (increasing) - Between 0 and $$\sqrt{\frac{3}{2}}$$, pick $$x = 1$$: $$f'(1) = 4(1) - 6(1) = 4 - 6 = -2 < 0$$ (decreasing) - For $$x > \sqrt{\frac{3}{2}}$$, pick $$x = 2$$: $$f'(2) = 4(8) - 6(2) = 32 - 12 = 20 > 0$$ (increasing) 12. **Turning points:** At critical points where $$f'$$ changes sign: - $$x = -\sqrt{\frac{3}{2}}$$: from decreasing to increasing (local min) - $$x = 0$$: from increasing to decreasing (local max) - $$x = \sqrt{\frac{3}{2}}$$: from decreasing to increasing (local min) 13. **Second derivative:** $$f''(x) = 12x^2 - 6$$ 14. **Possible points of inflection:** Solve $$f''(x) = 0$$: $$12x^2 - 6 = 0 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}}$$ 15. **Concavity test:** - For $$x < -\frac{1}{\sqrt{2}}$$, pick $$x = -1$$: $$f''(-1) = 12(1) - 6 = 6 > 0$$ (concave up) - Between $$-\frac{1}{\sqrt{2}}$$ and $$\frac{1}{\sqrt{2}}$$, pick $$x = 0$$: $$f''(0) = -6 < 0$$ (concave down) - For $$x > \frac{1}{\sqrt{2}}$$, pick $$x = 1$$: $$f''(1) = 6 > 0$$ (concave up) 16. **Actual points of inflection:** At $$x = \pm \frac{1}{\sqrt{2}}$$ where concavity changes. 17. **Symmetry:** Since $$f(x)$$ contains only even powers, $$f(-x) = f(x)$$, so $$f$$ is even and symmetric about the y-axis. 18. **End behavior:** Leading term $$x^4$$ dominates. - As $$x \to \infty$$, $$f(x) \to \infty$$ - As $$x \to -\infty$$, $$f(x) \to \infty$$ 19. **Max/min on interval [0,2]:** Evaluate $$f$$ at critical points in [0,2] and endpoints: - $$f(0) = 2$$ - $$f(\sqrt{\frac{3}{2}}) = (\sqrt{\frac{3}{2}})^4 - 3(\sqrt{\frac{3}{2}})^2 + 2 = \left(\frac{3}{2}\right)^2 - 3 \cdot \frac{3}{2} + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9}{4} - \frac{18}{4} + \frac{8}{4} = -\frac{1}{4}$$ - $$f(2) = 16 - 12 + 2 = 6$$ So on [0,2], minimum is approximately $$-0.25$$ at $$x = \sqrt{\frac{3}{2}}$$, maximum is 6 at $$x=2$$. 20. **Rolle's theorem on [-2,2]:** Check $$f(-2)$$ and $$f(2)$$: $$f(-2) = 16 - 12 + 2 = 6$$ $$f(2) = 6$$ Since $$f(-2) = f(2)$$ and $$f$$ is continuous and differentiable on $$[-2,2]$$, Rolle's theorem applies. Find $$c$$ in $$(-2,2)$$ such that $$f'(c) = 0$$. From step 10, critical numbers are $$x=0, \pm \sqrt{\frac{3}{2}}$$ all in $$(-2,2)$$. **Final answers:** - Domain: $$(-\infty, \infty)$$ - Range: $$[-\frac{1}{4}, \infty)$$ - Vertical asymptotes: None - Zeros: $$\pm 1, \pm \sqrt{2}$$ - Multiplicity: 1 for each zero - Factored form: $$(x-1)(x+1)(x-\sqrt{2})(x+\sqrt{2})$$ - Y-intercept: 2 - First derivative: $$4x^3 - 6x$$ - Critical numbers: $$0, \pm \sqrt{\frac{3}{2}}$$ - Increasing on $$(-\sqrt{\frac{3}{2}}, 0) \cup (\sqrt{\frac{3}{2}}, \infty)$$ - Decreasing on $$(-\infty, -\sqrt{\frac{3}{2}}) \cup (0, \sqrt{\frac{3}{2}})$$ - Turning points: local min at $$\pm \sqrt{\frac{3}{2}}$$, local max at 0 - Second derivative: $$12x^2 - 6$$ - Points of inflection: $$\pm \frac{1}{\sqrt{2}}$$ - Concave up on $$(-\infty, -\frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, \infty)$$ - Concave down on $$(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$$ - Symmetry: even function - Max/min on [0,2]: min approx $$-0.25$$ at $$x=\sqrt{\frac{3}{2}}$$, max 6 at $$x=2$$ - Rolle's theorem applies on $$[-2,2]$$ with $$c = 0, \pm \sqrt{\frac{3}{2}}$$ where $$f'(c) = 0$$.