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To determine where the function is increasing or decreasing, we need to find its derivative, $f'(x)$. We use the quotient rule: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
Here, $u(x) = x - 4$ and $v(x) = x + 1$. Thus, $u'(x) = 1$ and $v'(x) = 1$. Applying the quotient rule, we get: $$f'(x) = \frac{1(x+1) - (x-4)(1)}{(x+1)^2} = \frac{x+1 - x + 4}{(x+1)^2} = \frac{5}{(x+1)^2}$$
Since $(x+1)^2$ is always positive for $x \ne -1$, and 5 is positive, $f'(x) = \frac{5}{(x+1)^2}$ is always positive for all $x \ne -1$.
Since $f'(x) > 0$ for all $x \ne -1$, the function $f(x)$ is increasing on the intervals $(-\infty, -1)$ and $(-1, \infty)$. The function is never decreasing.
Final Answer: Increasing on $(-\infty, -1)$ and $(-1, \infty)$. Never decreasing.
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