1. **Problem statement:** Given the function $g(x) = \frac{2x+3}{x+1}$, find its domain, parity, limits at domain boundaries, asymptotes, first and second derivatives with sign tables, intercepts, points of intersection with asymptotes, other points, max and min points, intervals of increase/decrease, concavity, and inflection points. 2. **Domain:** The function is defined where the denominator is not zero. $$x+1 \neq 0 \implies x \neq -1$$ So, the domain is $\mathbb{R} \setminus \{-1\}$. 3. **Parity:** Check if $g(-x) = g(x)$ (even) or $g(-x) = -g(x)$ (odd). $$g(-x) = \frac{2(-x)+3}{-x+1} = \frac{-2x+3}{-x+1}$$ This is not equal to $g(x)$ or $-g(x)$, so $g$ is neither even nor odd. 4. **Limits at domain boundaries:** - As $x \to -1^-$: $$\lim_{x \to -1^-} g(x) = \lim_{x \to -1^-} \frac{2x+3}{x+1}$$ Denominator approaches $0^-$, numerator approaches $2(-1)+3=1$, so limit is $-\infty$. - As $x \to -1^+$: Denominator approaches $0^+$, numerator $1$, so limit is $+\infty$. - As $x \to \pm \infty$: $$\lim_{x \to \pm \infty} \frac{2x+3}{x+1} = \lim_{x \to \pm \infty} \frac{2 + \frac{3}{x}}{1 + \frac{1}{x}} = 2$$ 5. **Asymptotes:** - Vertical asymptote at $x = -1$. - Horizontal asymptote at $y = 2$. 6. **First derivative:** Using quotient rule: $$g'(x) = \frac{(2)(x+1) - (2x+3)(1)}{(x+1)^2} = \frac{2x + 2 - 2x - 3}{(x+1)^2} = \frac{-1}{(x+1)^2}$$ Since denominator squared is positive, $g'(x) < 0$ for all $x \neq -1$. 7. **Second derivative:** $$g''(x) = \frac{d}{dx} \left( \frac{-1}{(x+1)^2} \right) = -1 \cdot \frac{d}{dx} (x+1)^{-2} = -1 \cdot (-2)(x+1)^{-3} = \frac{2}{(x+1)^3}$$ 8. **Sign tables:** - $g'(x)$ is always negative, so $g$ is strictly decreasing on each interval $(-\infty, -1)$ and $(-1, \infty)$. - $g''(x)$ sign depends on $(x+1)^3$: - For $x < -1$, $(x+1)^3 < 0$, so $g''(x) < 0$ (concave down). - For $x > -1$, $(x+1)^3 > 0$, so $g''(x) > 0$ (concave up). 9. **Intercepts:** - $y$-intercept at $x=0$: $$g(0) = \frac{0+3}{0+1} = 3$$ - $x$-intercept where numerator zero: $$2x + 3 = 0 \implies x = -\frac{3}{2}$$ So intercepts at $(0,3)$ and $(-\frac{3}{2}, 0)$. 10. **Points of intersection with asymptotes:** - Vertical asymptote $x=-1$ is not in domain, so no intersection. - Horizontal asymptote $y=2$: Set $g(x) = 2$: $$\frac{2x+3}{x+1} = 2 \implies 2x + 3 = 2x + 2 \implies 3 = 2$$ No solution, so no intersection with horizontal asymptote. 11. **Other points:** Any point in domain can be chosen; for example, $x=1$: $$g(1) = \frac{2(1)+3}{1+1} = \frac{5}{2} = 2.5$$ 12. **Max and min points:** Since $g'(x) < 0$ everywhere, no local maxima or minima. 13. **Increasing and decreasing intervals:** $g$ is strictly decreasing on $(-\infty, -1)$ and $(-1, \infty)$. 14. **Concavity and inflection points:** Concave down on $(-\infty, -1)$, concave up on $(-1, \infty)$. At $x=-1$, function undefined, so no inflection point. **Final answers:** - Domain: $\mathbb{R} \setminus \{-1\}$ - Parity: neither even nor odd - Limits: $\lim_{x \to -1^-} g(x) = -\infty$, $\lim_{x \to -1^+} g(x) = +\infty$, $\lim_{x \to \pm \infty} g(x) = 2$ - Asymptotes: vertical at $x=-1$, horizontal at $y=2$ - $g'(x) = \frac{-1}{(x+1)^2} < 0$ always - $g''(x) = \frac{2}{(x+1)^3}$, negative for $x<-1$, positive for $x>-1$ - Intercepts: $(0,3)$ and $(-\frac{3}{2},0)$ - No intersection with asymptotes - No max or min points - Decreasing on $(-\infty, -1)$ and $(-1, \infty)$ - Concave down on $(-\infty, -1)$, concave up on $(-1, \infty)$ - No inflection points