1. **Statement of the problem:** We have functions: $$g(x)=x^3+6x-4$$ $$f(x)=\frac{x^3+2}{x^2+2}$$ and $$h(x)=-|f(x)|$$ Given an interval $0.6<\alpha<0.7$, and that $f'(x)=\frac{x\cdot g(x)}{(x^2+2)^2}$. We need to: - Find $\lim_{x\to +\infty} f(x)$ and $\lim_{x\to -\infty} f(x)$. - Verify that $\frac{3}{2}\alpha = f'(\alpha)$. - Locate the root $x_0$ of $f(x)=0$ between $-1.3$ and $-1.2$. - Discuss function $h(x)$ 2. **Limits of $f(x)$ at infinity:** The function is a rational function where degrees of numerator and denominator are 3 and 2 respectively. For large $x$, the highest power terms dominate: $$f(x) \approx \frac{x^3}{x^2} = x$$ Thus: - As $x \to +\infty$, $f(x) \to +\infty$. - As $x \to -\infty$, $f(x) \to -\infty$. **Answer:** $$\lim_{x \to +\infty} f(x) = +\infty$$ $$\lim_{x \to -\infty} f(x) = -\infty$$ 3. **Verifying $\frac{3}{2} \alpha = f'(\alpha)$ where $f'(x)=\frac{x g(x)}{(x^2+2)^2}$:** Substitute $f'(\alpha) = \frac{3}{2} \alpha$: $$\frac{x g(x)}{(x^2+2)^2} = \frac{3}{2} x$$ Setting $x=\alpha$: $$\frac{\alpha g(\alpha)}{(\alpha^2+2)^2} = \frac{3}{2} \alpha$$ Divide both sides by $\alpha$ (nonzero): $$\frac{g(\alpha)}{(\alpha^2+2)^2} = \frac{3}{2}$$ Multiply both sides: $$g(\alpha) = \frac{3}{2} (\alpha^2+2)^2$$ Recall $g(x)=x^3+6x-4$, so: $$\alpha^3 + 6\alpha -4 = \frac{3}{2} (\alpha^2 + 2)^2$$ This equation can be checked numerically for $0.6 < \alpha < 0.7$. 4. **Locating root $x_0$ of $f(x) = 0$ between $-1.3$ and $-1.2$:** We have: $$f(x) = \frac{x^3 + 2}{x^2 + 2} = 0 \implies x^3 + 2 = 0$$ So: $$x^3 = -2 \implies x = -\sqrt[3]{2} \approx -1.26$$ Since $-1.3 < -1.26 < -1.2$, the root $x_0$ lies exactly in that interval. 5. **Function $h(x) = -|f(x)|$ and its graph:** - $h(x)$ is the negative absolute value of $f(x)$. - This means $h(x) \leq 0$ for all $x$. - The graph of $h$ is a reflection of the positive part of $|f(x)|$ below the $x$-axis. **Summary:** - $\lim_{x \to +\infty} f(x) = +\infty$ - $\lim_{x \to -\infty} f(x) = -\infty$ - $f'(x) = \frac{x g(x)}{(x^2+2)^2}$ and satisfies $\frac{3}{2} \alpha = f'(\alpha)$ yielding $g(\alpha) = \frac{3}{2} (\alpha^2+2)^2$ - Root $x_0$ of $f(x)=0$ is approximately $-1.26$, located between $-1.3$ and $-1.2$ - $h(x)$ is negative absolute value of $f(x)$ reflecting $f$ downwards