1. **State the problem:** Find the intervals where the function $f(x) = (x-2)(x+3)^2(x+7)$ is increasing or decreasing. 2. **Find the derivative:** To determine increasing or decreasing intervals, we find $f'(x)$ and analyze its sign. Use the product rule and chain rule: $$f(x) = (x-2)(x+3)^2(x+7)$$ Let $u = (x-2)$, $v = (x+3)^2$, and $w = (x+7)$. Then, $$f'(x) = u'vw + uv'w + uvw'$$ Calculate each derivative: $$u' = 1$$ $$v' = 2(x+3)$$ $$w' = 1$$ So, $$f'(x) = 1 \cdot (x+3)^2 \cdot (x+7) + (x-2) \cdot 2(x+3) \cdot (x+7) + (x-2)(x+3)^2 \cdot 1$$ 3. **Simplify $f'(x)$:** $$f'(x) = (x+3)^2(x+7) + 2(x-2)(x+3)(x+7) + (x-2)(x+3)^2$$ Factor where possible: Group terms: $$= (x+3)(x+7)[(x+3) + 2(x-2)] + (x-2)(x+3)^2$$ Calculate inside the bracket: $$ (x+3) + 2(x-2) = x+3 + 2x -4 = 3x -1$$ So, $$f'(x) = (x+3)(x+7)(3x -1) + (x-2)(x+3)^2$$ Expand the second term: $$ (x-2)(x+3)^2 = (x-2)(x^2 + 6x + 9) = x^3 + 6x^2 + 9x - 2x^2 - 12x - 18 = x^3 + 4x^2 - 3x - 18$$ Expand the first term: $$(x+3)(x+7)(3x -1)$$ First, $(x+3)(x+7) = x^2 + 10x + 21$ Then multiply by $(3x -1)$: $$ (x^2 + 10x + 21)(3x -1) = 3x^3 - x^2 + 30x^2 - 10x + 63x - 21 = 3x^3 + 29x^2 + 53x - 21$$ Add both parts: $$f'(x) = (3x^3 + 29x^2 + 53x - 21) + (x^3 + 4x^2 - 3x - 18) = 4x^3 + 33x^2 + 50x - 39$$ 4. **Find critical points:** Solve $f'(x) = 0$: $$4x^3 + 33x^2 + 50x - 39 = 0$$ Try rational roots using factors of 39 and 4. Test $x=1$: $$4(1)^3 + 33(1)^2 + 50(1) - 39 = 4 + 33 + 50 - 39 = 48 \neq 0$$ Test $x=-3$: $$4(-3)^3 + 33(-3)^2 + 50(-3) - 39 = 4(-27) + 33(9) - 150 - 39 = -108 + 297 - 150 - 39 = 0$$ So, $x = -3$ is a root. Divide polynomial by $(x+3)$: Using synthetic division: Coefficients: 4 | 33 | 50 | -39 Bring down 4. Multiply by -3: -12, add to 33: 21. Multiply by -3: -63, add to 50: -13. Multiply by -3: 39, add to -39: 0. Quotient: $4x^2 + 21x - 13$ Solve quadratic: $$4x^2 + 21x - 13 = 0$$ Use quadratic formula: $$x = \frac{-21 \pm \sqrt{21^2 - 4 \cdot 4 \cdot (-13)}}{2 \cdot 4} = \frac{-21 \pm \sqrt{441 + 208}}{8} = \frac{-21 \pm \sqrt{649}}{8}$$ Approximate roots: $$\sqrt{649} \approx 25.48$$ So, $$x_1 = \frac{-21 - 25.48}{8} = \frac{-46.48}{8} \approx -5.81$$ $$x_2 = \frac{-21 + 25.48}{8} = \frac{4.48}{8} \approx 0.56$$ 5. **Determine intervals of increase/decrease:** Critical points: $x \approx -5.81, -3, 0.56$ Test intervals: - For $x < -5.81$, pick $x = -6$: $$f'(-6) = 4(-6)^3 + 33(-6)^2 + 50(-6) - 39 = -864 + 1188 - 300 - 39 = -15 < 0$$ - For $-5.81 < x < -3$, pick $x = -4$: $$f'(-4) = 4(-64) + 33(16) + 50(-4) - 39 = -256 + 528 - 200 - 39 = 33 > 0$$ - For $-3 < x < 0.56$, pick $x = 0$: $$f'(0) = -39 < 0$$ - For $x > 0.56$, pick $x = 1$: $$f'(1) = 4 + 33 + 50 - 39 = 48 > 0$$ 6. **Conclusion:** - $f(x)$ is decreasing on $(-\infty, -5.81)$ - Increasing on $(-5.81, -3)$ - Decreasing on $(-3, 0.56)$ - Increasing on $(0.56, \infty)$ **Final answer:** $$\text{Increasing intervals: } (-5.81, -3) \cup (0.56, \infty)$$ $$\text{Decreasing intervals: } (-\infty, -5.81) \cup (-3, 0.56)$$