The local maximum value l and local minimum v | vedclass

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(A) Given $f(x) = \frac{x^2+2x+2}{x+1} = \frac{(x+1)^2+1}{x+1} = (x+1) + \frac{1}{x+1}$.Let $u = x+1$. Then $f(u) = u + \frac{1}{u}$.To find local ext

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(A) Given $f(x) = \frac{x^2+2x+2}{x+1} = \frac{(x+1)^2+1}{x+1} = (x+1) + \frac{1}{x+1}$.Let $u = x+1$. Then $f(u) = u + \frac{1}{u}$.To find local extrema, we find the derivative: $f'(u) = 1 - \frac{1}{u^2}$.Setting $f'(u) = 0$, we get $u^2 = 1$, so $u = 1$ or $u = -1$.For $u = 1$, $x+1 = 1 \implies x = 0$. $f(0) = 0 + \frac{1}{1} = 2$. This is the local minimum value $m = 2$ at $\beta = 0$.For $u = -1$, $x+1 = -1 \implies x = -2$. $f(-2) = -2 + \frac{1}{-1} = -2 - 1 = -3$. This is the local maximum value $l = -3$ at $\alpha = -2$.Thus, $l = -3, m = 2, \alpha = -2, \beta = 0$.The value of $\frac{l+m}{\alpha+\beta} = \frac{-3+2}{-2+0} = \frac{-1}{-2} = \frac{1}{2}$.Wait, checking the options, let us re-evaluate. If $f(x) = x+1 + \frac{1}{x+1}$, for $x > -1$, $u > 0$, $f(u) \ge 2$ ($AM$-$GM$). For $x < -1$, $u < 0$, $f(u) \le -2$. The local maximum is $-2$ at $x = -2$ and local minimum is $2$ at $x = 0$. The expression $\frac{l+m}{\alpha+\beta} = \frac{-2+2}{-2+0} = 0$.

The local maximum value $l$ and local minimum value $m$ of $f(x) = \frac{x^2+2x+2}{x+1}$ in $R - \{-1\}$ exist at $\alpha, \beta$ respectively, then $\frac{l+m}{\alpha+\beta} =$

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