1. **Problem statement:** We study the function $f(x) = x(\sqrt{x} - 2)^2$ for $x > 0$. 2. **Limits at infinity:** - Compute $\lim_{x \to +\infty} f(x)$. - Compute $\lim_{x \to +\infty} \frac{f(x)}{x}$. 3. **Derivative given:** $f'(x) = 2(\sqrt{x} - 1)(\sqrt{x} - 2)$. 4. **Concavity and inflection point:** We study the concavity of $f$ by finding $f''(x)$ and determine the unique inflection point $I$. --- **Step 1: Simplify $f(x)$** Recall $f(x) = x(\sqrt{x} - 2)^2 = x(x^{1/2} - 2)^2$. Expand the square: $$ (\sqrt{x} - 2)^2 = x - 4\sqrt{x} + 4 $$ So, $$ f(x) = x(x - 4\sqrt{x} + 4) = x^2 - 4x^{3/2} + 4x $$ --- **Step 2: Compute $\lim_{x \to +\infty} f(x)$** As $x \to +\infty$, the dominant term is $x^2$, so $$ \lim_{x \to +\infty} f(x) = +\infty $$ --- **Step 3: Compute $\lim_{x \to +\infty} \frac{f(x)}{x}$** Divide $f(x)$ by $x$: $$ \frac{f(x)}{x} = \frac{x^2 - 4x^{3/2} + 4x}{x} = x - 4x^{1/2} + 4 $$ As $x \to +\infty$, $x$ dominates, so $$ \lim_{x \to +\infty} \frac{f(x)}{x} = +\infty $$ --- **Step 4: Verify $f'(x)$ and find $f''(x)$** Given: $$ f'(x) = 2(\sqrt{x} - 1)(\sqrt{x} - 2) $$ Rewrite $f'(x)$ in terms of $x^{1/2}$: $$ f'(x) = 2(x^{1/2} - 1)(x^{1/2} - 2) $$ Expand: $$ f'(x) = 2(x - 3x^{1/2} + 2) = 2x - 6x^{1/2} + 4 $$ Differentiate again to find $f''(x)$: Recall derivative rules: $$ \frac{d}{dx} x = 1, \quad \frac{d}{dx} x^{1/2} = \frac{1}{2} x^{-1/2} $$ So, $$ f''(x) = 2 - 6 \times \frac{1}{2} x^{-1/2} + 0 = 2 - 3 x^{-1/2} = 2 - \frac{3}{\sqrt{x}} $$ --- **Step 5: Find inflection point $I$ where $f''(x) = 0$** Set: $$ 2 - \frac{3}{\sqrt{x}} = 0 $$ Solve for $x$: $$ 2 = \frac{3}{\sqrt{x}} \implies \sqrt{x} = \frac{3}{2} \implies x = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25 $$ --- **Step 6: Find $f(x)$ at $x = \frac{9}{4}$** Calculate $\sqrt{x} = \frac{3}{2}$. Recall: $$ f(x) = x(\sqrt{x} - 2)^2 $$ So, $$ f\left(\frac{9}{4}\right) = \frac{9}{4} \left(\frac{3}{2} - 2\right)^2 = \frac{9}{4} \left(-\frac{1}{2}\right)^2 = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16} = 0.5625 $$ --- **Step 7: Concavity analysis** - For $x > \frac{9}{4}$, $f''(x) = 2 - \frac{3}{\sqrt{x}} > 0$ so $f$ is concave up. - For $0 < x < \frac{9}{4}$, $f''(x) < 0$ so $f$ is concave down. Thus, there is a unique inflection point $I$ at $$ I \left(\frac{9}{4}, \frac{9}{16}\right) $$ --- **Final answers:** - $\lim_{x \to +\infty} f(x) = +\infty$ - $\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty$ - Unique inflection point $I$ at $\left(\frac{9}{4}, \frac{9}{16}\right)$ ---