1. **State the problem:** Given the function $$x = x^3 + 3x^2 - 9x + 6$$, we need to find the first derivative, determine critical points, identify intervals of increase/decrease and concavity, and plot the function on the domain $$[-5,5]$$ and range $$[-2,10]$$. 2. **Find the first derivative:** The derivative of $$f(x) = x^3 + 3x^2 - 9x + 6$$ is found using the power rule: $$f'(x) = 3x^2 + 6x - 9$$ 3. **Determine critical points:** Critical points occur where $$f'(x) = 0$$ or $$f'(x)$$ is undefined. Since $$f'(x)$$ is a polynomial, it is defined everywhere. Solve: $$3x^2 + 6x - 9 = 0$$ Divide both sides by 3: $$\cancel{3}x^2 + \cancel{3}2x - \cancel{3}3 = 0 \Rightarrow x^2 + 2x - 3 = 0$$ Factor: $$(x + 3)(x - 1) = 0$$ So, $$x = -3 \quad \text{or} \quad x = 1$$ 4. **Identify intervals of increase/decrease:** Test values in intervals determined by critical points: - For $$x < -3$$, pick $$x = -4$$: $$f'(-4) = 3(-4)^2 + 6(-4) - 9 = 48 - 24 - 9 = 15 > 0$$ (increasing) - For $$-3 < x < 1$$, pick $$x = 0$$: $$f'(0) = -9 < 0$$ (decreasing) - For $$x > 1$$, pick $$x = 2$$: $$f'(2) = 3(4) + 12 - 9 = 15 > 0$$ (increasing) 5. **Identify concavity:** Find the second derivative: $$f''(x) = 6x + 6$$ Set $$f''(x) = 0$$ to find inflection points: $$6x + 6 = 0 \Rightarrow x = -1$$ Test concavity: - For $$x < -1$$, pick $$x = -2$$: $$f''(-2) = -12 + 6 = -6 < 0$$ (concave down) - For $$x > -1$$, pick $$x = 0$$: $$f''(0) = 6 > 0$$ (concave up) 6. **Plot the function:** Use domain $$[-5,5]$$ and range $$[-2,10]$$. 7. **Commands in Maple (or similar software):** - Derivative: `diff(x^3 + 3*x^2 - 9*x + 6, x);` - Solve derivative zero: `solve(3*x^2 + 6*x - 9 = 0, x);` - Second derivative: `diff(diff(x^3 + 3*x^2 - 9*x + 6, x), x);` - Plot: `plot(x^3 + 3*x^2 - 9*x + 6, x = -5..5, y = -2..10);` **Final answers:** - First derivative: $$f'(x) = 3x^2 + 6x - 9$$ - Critical points: $$x = -3, 1$$ - Increasing on $$(-\infty, -3) \cup (1, \infty)$$, decreasing on $$(-3, 1)$$ - Concave down on $$(-\infty, -1)$$, concave up on $$(-1, \infty)$$