1. **Stating the problem:** We are given the function $$f(x) = (2 - x^2)(\ln x + 1)$$ and asked to analyze it according to the following points: 2. **Domain (תחום הגדרה):** - The term $$\ln x$$ is defined only for $$x > 0$$. - Therefore, the domain of $$f(x)$$ is $$x > 0$$. 3. **Vertical asymptote (אסימפטוטה אנכית):** - Since $$\ln x$$ tends to $$-\infty$$ as $$x \to 0^+$$, and $$2 - x^2$$ tends to 2 near 0, the function $$f(x)$$ tends to $$2 \cdot (-\infty) = -\infty$$. - Thus, there is a vertical asymptote at $$x = 0$$. 4. **Behavior near the asymptote (זינוק עם הפונקציה):** - As $$x \to 0^+$$, $$f(x) \to -\infty$$. - As $$x \to +\infty$$, $$2 - x^2 \to -\infty$$ and $$\ln x + 1 \to +\infty$$. - The product of a large negative and large positive tends to $$-\infty$$. 5. **Critical points and extrema (נקודות קיצון ונעילה):** - To find critical points, compute $$f'(x)$$: $$f(x) = (2 - x^2)(\ln x + 1)$$ Using product rule: $$f'(x) = (2 - x^2) \cdot \frac{1}{x} + (\ln x + 1) \cdot (-2x)$$ Simplify: $$f'(x) = \frac{2 - x^2}{x} - 2x(\ln x + 1) = \frac{2}{x} - x - 2x \ln x - 2x$$ Combine terms: $$f'(x) = \frac{2}{x} - 3x - 2x \ln x$$ Set $$f'(x) = 0$$: $$\frac{2}{x} - 3x - 2x \ln x = 0$$ Multiply both sides by $$x$$: $$2 - 3x^2 - 2x^2 \ln x = 0$$ Rearranged: $$2 = 3x^2 + 2x^2 \ln x = x^2 (3 + 2 \ln x)$$ So: $$x^2 (3 + 2 \ln x) = 2$$ This transcendental equation can be solved numerically to find critical points. 6. **Sketch of $$f(x)$$ (עיר סקיצה של f(x))**: - Domain: $$x > 0$$. - Vertical asymptote at $$x=0$$ with $$f(x) \to -\infty$$. - For large $$x$$, $$f(x) \to -\infty$$. - Critical points found by solving $$x^2 (3 + 2 \ln x) = 2$$. 7. **Check the equation of $$g(x)$$ (בדיקה את המשוואה הנכונה של g(x))**: Given: $$g(x) = -2 - f(x)$$ This means $$g(x)$$ is the function $$f(x)$$ reflected and shifted down by 2. **Final answer:** - Domain: $$x > 0$$. - Vertical asymptote at $$x=0$$. - Critical points satisfy $$x^2 (3 + 2 \ln x) = 2$$. - $$g(x) = -2 - f(x)$$.