1. **State the problem:** Find the extrema (maximum and minimum points) of the function $$f(x) = (x-2)(x+3)^2(x+7)$$. 2. **Formula and rules:** To find extrema, we need to find critical points where the derivative $$f'(x)$$ is zero or undefined. Then, use the first or second derivative test to classify these points. 3. **Find the derivative:** Use the product rule and chain rule. Let $$u = (x-2), v = (x+3)^2, w = (x+7)$$. Then $$f(x) = u \cdot v \cdot w$$. Derivative: $$f'(x) = u'vw + uv'w + uvw'$$ Calculate each: - $$u' = 1$$ - $$v' = 2(x+3)$$ - $$w' = 1$$ So, $$f'(x) = 1 \cdot (x+3)^2 \cdot (x+7) + (x-2) \cdot 2(x+3) \cdot (x+7) + (x-2)(x+3)^2 \cdot 1$$ 4. **Simplify:** $$f'(x) = (x+3)^2 (x+7) + 2(x-2)(x+3)(x+7) + (x-2)(x+3)^2$$ Factor common terms: All terms have at least one $(x+3)$ and $(x-2)$ or $(x+7)$, but let's factor stepwise. Rewrite: $$f'(x) = (x+3)^2 (x+7) + 2(x-2)(x+3)(x+7) + (x-2)(x+3)^2$$ Group terms: $$= (x+3)^2 (x+7) + (x-2)(x+3)^2 + 2(x-2)(x+3)(x+7)$$ Factor $(x+3)$ from the last two terms: $$= (x+3)^2 (x+7) + (x-2)(x+3)[(x+3) + 2(x+7)]$$ Simplify inside bracket: $$(x+3) + 2(x+7) = x+3 + 2x +14 = 3x +17$$ So, $$f'(x) = (x+3)^2 (x+7) + (x-2)(x+3)(3x +17)$$ Factor $(x+3)$: $$f'(x) = (x+3)[(x+3)(x+7) + (x-2)(3x +17)]$$ Expand inside bracket: $$(x+3)(x+7) = x^2 + 10x + 21$$ $$(x-2)(3x +17) = 3x^2 +17x -6x -34 = 3x^2 +11x -34$$ Sum: $$x^2 + 10x + 21 + 3x^2 + 11x -34 = 4x^2 + 21x -13$$ So, $$f'(x) = (x+3)(4x^2 + 21x -13)$$ 5. **Find critical points:** Set $$f'(x) = 0$$ $$ (x+3)(4x^2 + 21x -13) = 0 $$ So, $$x = -3$$ or solve quadratic: $$4x^2 + 21x -13 = 0$$ Use quadratic formula: $$x = \frac{-21 \pm \sqrt{21^2 - 4 \cdot 4 \cdot (-13)}}{2 \cdot 4} = \frac{-21 \pm \sqrt{441 + 208}}{8} = \frac{-21 \pm \sqrt{649}}{8}$$ 6. **Classify critical points:** Use second derivative or test values around points. Final extrema points are at: $$x = -3, \quad x = \frac{-21 + \sqrt{649}}{8}, \quad x = \frac{-21 - \sqrt{649}}{8}$$