1. **State the problem:** We have a piecewise function: $$f(x) = \begin{cases}(x - 1) \sqrt[3]{x^2} & x \leq 0 \\ x^2 \arctan\left(\frac{1}{x}\right) & x > 0\end{cases}$$ We need to determine the domain $D_f$, calculate limits, show continuity, study differentiability at 0, prove an inequality using the Mean Value Theorem, and deduce the asymptote at $+\infty$. 2. **Determine the domain $D_f$:** - For $x \leq 0$, $f(x) = (x-1) \sqrt[3]{x^2}$ is defined for all real $x$ since cube root is defined everywhere. - For $x > 0$, $f(x) = x^2 \arctan(1/x)$ is defined for all $x > 0$ since $1/x$ is defined and $\arctan$ is defined everywhere. Thus, the domain is $D_f = \mathbb{R} \setminus \{0\}$ plus we check if $f$ is defined at 0 from the left side (since $x=0$ is included in $x \leq 0$ part). The cube root is defined at 0, so $f(0) = (0-1) \sqrt[3]{0} = 0$. Therefore, $D_f = \mathbb{R}$. 3. **Calculate limits at 0:** - Left limit as $x \to 0^-$: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x-1) \sqrt[3]{x^2} = (0-1) \cdot 0 = 0$$ - Right limit as $x \to 0^+$: Since $\arctan(1/x) \to \arctan(+\infty) = \frac{\pi}{2}$, $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 \cdot \frac{\pi}{2} = 0$$ Limits from both sides equal 0. 4. **Show continuity on $D_f$:** Since $f$ is defined piecewise with continuous expressions on $(-\infty,0]$ and $(0,\infty)$ and the limits at 0 coincide with $f(0)$, $f$ is continuous on $\mathbb{R}$. 5. **Study differentiability at 0:** - Compute left derivative: $$f'_-(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(h-1) \sqrt[3]{h^2} - 0}{h} = \lim_{h \to 0^-} \frac{(h-1) |h|^{2/3}}{h}$$ Since $h<0$, $|h| = -h$, so: $$= \lim_{h \to 0^-} \frac{(h-1)(-h)^{2/3}}{h} = \lim_{h \to 0^-} (h-1)(-h)^{-1/3}$$ As $h \to 0^-$, $h-1 \to -1$ and $(-h)^{-1/3} \to +\infty$, so the limit diverges to $-\infty$. - Compute right derivative: $$f'_+(0) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 \arctan(1/h)}{h} = \lim_{h \to 0^+} h \arctan(1/h)$$ Since $\arctan(1/h) \to \frac{\pi}{2}$, $$= \lim_{h \to 0^+} h \cdot \frac{\pi}{2} = 0$$ Since left and right derivatives differ, $f$ is not differentiable at 0. 6. **Show by Mean Value Theorem (MVT) that for all $t > 0$:** $$0 < \frac{t - \arctan t}{t} < t^2$$ - Define $g(t) = t - \arctan t$. - $g(0) = 0$. - By MVT, for $t > 0$, there exists $c \in (0,t)$ such that: $$g'(c) = \frac{g(t) - g(0)}{t - 0} = \frac{t - \arctan t}{t}$$ - Compute derivative: $$g'(x) = 1 - \frac{1}{1+x^2} = \frac{x^2}{1+x^2}$$ - Since $x^2/(1+x^2) > 0$ for $x > 0$, we have: $$0 < \frac{t - \arctan t}{t} = g'(c) = \frac{c^2}{1+c^2} < c^2 < t^2$$ 7. **Deduce asymptote $y = x$ of $C_f$ at $+\infty$:** - For $x > 0$, $$f(x) = x^2 \arctan\left(\frac{1}{x}\right)$$ - As $x \to +\infty$, $\frac{1}{x} \to 0$, so $$\arctan\left(\frac{1}{x}\right) \sim \frac{1}{x}$$ - Thus, $$f(x) \sim x^2 \cdot \frac{1}{x} = x$$ - The difference: $$f(x) - x = x^2 \arctan\left(\frac{1}{x}\right) - x = x \left(x \arctan\left(\frac{1}{x}\right) - 1\right)$$ - Using the inequality from step 6 with $t = \frac{1}{x}$: $$0 < \frac{t - \arctan t}{t} < t^2 \implies 0 < 1 - \frac{\arctan t}{t} < t^2$$ - Substitute $t = \frac{1}{x}$: $$0 < 1 - x \arctan\left(\frac{1}{x}\right) < \frac{1}{x^2}$$ - Multiply by $x$: $$0 < x - x^2 \arctan\left(\frac{1}{x}\right) < \frac{1}{x}$$ - So, $$-\frac{1}{x} < f(x) - x < 0$$ - Hence, $$\lim_{x \to +\infty} (f(x) - x) = 0$$ Therefore, the line $y = x$ is an asymptote of $C_f$ at $+\infty$. **Final answers:** - Domain $D_f = \mathbb{R}$. - $f$ is continuous on $\mathbb{R}$. - $f$ is not differentiable at 0. - Inequality $0 < \frac{t - \arctan t}{t} < t^2$ holds for all $t > 0$. - $y = x$ is an asymptote of $C_f$ at $+\infty$.