1. **Problem Statement:** Given the function $f(x) = x^2 \ln x$ for $x > 0$, 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:** - The first derivative $f'(x)$ helps find critical points where $f'(x) = 0$ or undefined. - The second derivative $f''(x)$ helps determine 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. - The Second Derivative Test: - If $f'(c) = 0$ and $f''(c) > 0$, $f$ has a relative minimum at $x=c$. - If $f'(c) = 0$ and $f''(c) < 0$, $f$ has a relative maximum at $x=c$. 3. **Find the first derivative:** $$f(x) = x^2 \ln x$$ Using product rule: $$f'(x) = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x = x(2 \ln x + 1)$$ 4. **Find the second derivative:** $$f''(x) = \frac{d}{dx} \left[x(2 \ln x + 1)\right] = (1)(2 \ln x + 1) + x \cdot \frac{2}{x} = 2 \ln x + 1 + 2 = 2 \ln x + 3$$ 5. **Determine concavity:** - $f''(x) = 2 \ln x + 3$ - Set $f''(x) = 0$ to find possible inflection points: $$2 \ln x + 3 = 0 \implies \ln x = -\frac{3}{2} \implies x = e^{-3/2}$$ - For $x < e^{-3/2}$, since $\ln x < -\frac{3}{2}$, $f''(x) < 0$ so $f$ is concave down. - For $x > e^{-3/2}$, $f''(x) > 0$ so $f$ is concave up. 6. **Inflection point:** At $x = e^{-3/2}$, find $f(x)$: $$f\left(e^{-3/2}\right) = \left(e^{-3/2}\right)^2 \ln \left(e^{-3/2}\right) = e^{-3} \cdot \left(-\frac{3}{2}\right) = -\frac{3}{2} e^{-3}$$ So the inflection point is: $$\left(e^{-3/2}, -\frac{3}{2} e^{-3}\right)$$ 7. **Find critical numbers:** Set $f'(x) = 0$: $$x(2 \ln x + 1) = 0$$ Since $x > 0$, $x \neq 0$, so: $$2 \ln x + 1 = 0 \implies \ln x = -\frac{1}{2} \implies x = e^{-1/2}$$ 8. **Second Derivative Test at $x = e^{-1/2}$:** $$f''\left(e^{-1/2}\right) = 2 \ln \left(e^{-1/2}\right) + 3 = 2 \cdot \left(-\frac{1}{2}\right) + 3 = -1 + 3 = 2 > 0$$ Since $f''(e^{-1/2}) > 0$, there is a relative minimum at $x = e^{-1/2}$. 9. **Summary:** - (a) Concave down on $(0, e^{-3/2})$, concave up on $(e^{-3/2}, \infty)$. - (b) Inflection point at $\left(e^{-3/2}, -\frac{3}{2} e^{-3}\right)$. - (c) Critical number at $x = e^{-1/2}$ with a relative minimum there. --- **Final answers:** - Concave up on: $(e^{-3/2}, \infty)$ - Concave down on: $(0, e^{-3/2})$ - Inflection point: $\left(e^{-3/2}, -\frac{3}{2} e^{-3}\right)$ - Relative maxima at: (none) - Relative minima at: $e^{-1/2}$