1. **Problem Statement:** Given the function $f(x) = x^2 e^x$, we need to find: a) Intervals where $f$ is concave up and concave down. b) Inflection points. c) Critical numbers and relative extrema using the Second Derivative Test. 2. **Formulas and Rules:** - First derivative: $f'(x)$ gives critical points where $f'(x) = 0$ or undefined. - Second derivative: $f''(x)$ determines concavity. - If $f''(x) > 0$, $f$ is concave up. - If $f''(x) < 0$, $f$ is concave down. - Inflection points occur where $f''(x) = 0$ and concavity changes. - Second Derivative Test for extrema: - If $f'(c) = 0$ and $f''(c) > 0$, local minimum at $x=c$. - If $f'(c) = 0$ and $f''(c) < 0$, local maximum at $x=c$. 3. **Find derivatives:** - $f(x) = x^2 e^x$ - Use product rule: $(uv)' = u'v + uv'$ First derivative: $$f'(x) = (x^2)' e^x + x^2 (e^x)' = 2x e^x + x^2 e^x = e^x (2x + x^2) = e^x x (x + 2)$$ Second derivative: $$f''(x) = (f'(x))' = (e^x x (x + 2))'$$ Use product rule again: $$f''(x) = (e^x)' x (x+2) + e^x (x (x+2))' = e^x x (x+2) + e^x ( (x)(1) + (x+2)(1) )$$ Simplify derivative inside parentheses: $$ (x (x+2))' = (x^2 + 2x)' = 2x + 2$$ So: $$f''(x) = e^x x (x+2) + e^x (2x + 2) = e^x (x (x+2) + 2x + 2)$$ Expand $x(x+2)$: $$x^2 + 2x + 2x + 2 = x^2 + 4x + 2$$ Therefore: $$f''(x) = e^x (x^2 + 4x + 2)$$ 4. **Concavity:** Since $e^x > 0$ for all $x$, the sign of $f''(x)$ depends on $x^2 + 4x + 2$. Solve $x^2 + 4x + 2 = 0$ for inflection points: $$x = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2}$$ So inflection points at $x = -2 - \sqrt{2}$ and $x = -2 + \sqrt{2}$. Test intervals: - For $x < -2 - \sqrt{2}$, pick $x = -4$: $(-4)^2 + 4(-4) + 2 = 16 -16 + 2 = 2 > 0$ concave up. - Between $-2 - \sqrt{2}$ and $-2 + \sqrt{2}$, pick $x = -2$: $4 - 8 + 2 = -2 < 0$ concave down. - For $x > -2 + \sqrt{2}$, pick $x=0$: $0 + 0 + 2 = 2 > 0$ concave up. 5. **Inflection points coordinates:** Calculate $f(x)$ at inflection points: $$f(-2 \pm \sqrt{2}) = ((-2 \pm \sqrt{2})^2) e^{-2 \pm \sqrt{2}}$$ 6. **Critical numbers:** Solve $f'(x) = 0$: $$e^x x (x+2) = 0$$ Since $e^x \neq 0$, solve: $$x = 0 \quad \text{or} \quad x + 2 = 0 \Rightarrow x = -2$$ 7. **Second Derivative Test at critical points:** - At $x=0$: $$f''(0) = e^0 (0 + 0 + 2) = 1 \times 2 = 2 > 0$$ So local minimum at $x=0$. - At $x=-2$: $$f''(-2) = e^{-2} ((-2)^2 + 4(-2) + 2) = e^{-2} (4 - 8 + 2) = e^{-2} (-2) < 0$$ So local maximum at $x=-2$. **Final answers:** - Concave up on $(-\infty, -2 - \sqrt{2}) \cup (-2 + \sqrt{2}, \infty)$ - Concave down on $(-2 - \sqrt{2}, -2 + \sqrt{2})$ - Inflection points at $\left(-2 - \sqrt{2}, f(-2 - \sqrt{2})\right)$ and $\left(-2 + \sqrt{2}, f(-2 + \sqrt{2})\right)$ - Relative maxima at $x = -2$ - Relative minima at $x = 0$