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Given the function $f(x) = \frac{x-3}{x+1}$. Using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, where $u = x-3$ and $v = x+1$:
$$f'(x) = \frac{(1)(x+1) - (x-3)(1)}{(x+1)^2}$$ $$f'(x) = \frac{x+1-x+3}{(x+1)^2} = \frac{4}{(x+1)^2}$$A function is strictly increasing if $f'(x) > 0$. Here, $f'(x) = \frac{4}{(x+1)^2}$. Since the numerator is a positive constant ($4$) and the denominator is a square of a real number (which is always positive for $x \neq -1$), the derivative is always positive for all $x$ in the domain.
Since $f'(x) > 0$ for all $x \in \mathbb{R} \setminus \{-1\}$, the function is strictly increasing on its entire domain.
Final Answer: The function is increasing for all x in its domain, i.e., x \in (-\infty, -1) \cup (-1, \infty)
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