1. **State the problem:** We are given the function $f(x) = (x-1)^3 (x+1)$ and asked to use the second derivative test to find where $f(x)$ is concave up, concave down, and to locate all inflection points. 2. **Find the first derivative $f'(x)$:** Use the product rule for $f(x) = u(x)v(x)$ where $u = (x-1)^3$ and $v = (x+1)$. $$f'(x) = u'v + uv'$$ Calculate derivatives: $$u' = 3(x-1)^2, \quad v' = 1$$ So, $$f'(x) = 3(x-1)^2 (x+1) + (x-1)^3 (1)$$ 3. **Simplify $f'(x)$:** $$f'(x) = (x-1)^2 [3(x+1) + (x-1)] = (x-1)^2 (3x + 3 + x - 1) = (x-1)^2 (4x + 2)$$ 4. **Find the second derivative $f''(x)$:** Differentiate $f'(x) = (x-1)^2 (4x + 2)$ using product rule again. Let $a = (x-1)^2$, $b = 4x + 2$. $$a' = 2(x-1), \quad b' = 4$$ So, $$f''(x) = a'b + ab' = 2(x-1)(4x+2) + (x-1)^2 (4)$$ 5. **Simplify $f''(x)$:** $$f''(x) = 2(x-1)(4x+2) + 4(x-1)^2$$ Expand terms: $$2(x-1)(4x+2) = 2(4x^2 -4x + 2x - 2) = 2(4x^2 - 2x - 2) = 8x^2 - 4x - 4$$ $$4(x-1)^2 = 4(x^2 - 2x + 1) = 4x^2 - 8x + 4$$ Add them: $$f''(x) = (8x^2 - 4x - 4) + (4x^2 - 8x + 4) = 12x^2 - 12x + 0 = 12x^2 - 12x$$ 6. **Factor $f''(x)$:** $$f''(x) = 12x(x - 1)$$ 7. **Find inflection points:** Set $f''(x) = 0$. $$12x(x - 1) = 0 \implies x = 0 \text{ or } x = 1$$ 8. **Determine concavity intervals:** - For $x < 0$, choose $x = -1$: $f''(-1) = 12(-1)(-1 - 1) = 12(-1)(-2) = 24 > 0$ (concave up) - For $0 < x < 1$, choose $x = 0.5$: $f''(0.5) = 12(0.5)(0.5 - 1) = 12(0.5)(-0.5) = -3 < 0$ (concave down) - For $x > 1$, choose $x = 2$: $f''(2) = 12(2)(2 - 1) = 12(2)(1) = 24 > 0$ (concave up) 9. **Summary:** - $f(x)$ is concave up on $(-\infty, 0)$ and $(1, \infty)$ - $f(x)$ is concave down on $(0, 1)$ - Inflection points at $x = 0$ and $x = 1$ **Final answer:** $$\text{Concave up: } (-\infty, 0) \cup (1, \infty)$$ $$\text{Concave down: } (0, 1)$$ $$\text{Inflection points at } x = 0 \text{ and } x = 1$$