The function f(x) = frac{x - 2}{x + 1}, (x ne | vedclass

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(A) Given the function $f(x) = \frac{x - 2}{x + 1}$.To find the interval where the function is increasing,we calculate the derivative $f'(x)$ using th

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$( - \infty , -1) \cup (-1, \infty)$

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(A) Given the function $f(x) = \frac{x - 2}{x + 1}$.To find the interval where the function is increasing,we calculate the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(x + 1)(1) - (x - 2)(1)}{(x + 1)^2} = \frac{x + 1 - x + 2}{(x + 1)^2} = \frac{3}{(x + 1)^2}$.Since $(x + 1)^2 > 0$ for all $x 
eq -1$,it follows that $f'(x) = \frac{3}{(x + 1)^2} > 0$ for all $x$ in the domain $R \setminus \{-1\}$.Therefore,the function is increasing on its entire domain,which is $( - \infty , -1) \cup (-1, \infty)$.

The function $f(x) = \frac{x - 2}{x + 1}, (x \neq -1)$ is increasing on the interval:

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