Finding local extrema using the first derivative test

To simplify the analysis, we first rewrite $f(x)$ using polynomial division. Dividing $x^2 - 4$ by $x - 1$ yields: $f(x) = x + 1 + \dfrac{-3}{x-1}$. We can verify this: $(x+1)(x-1) + (-3) = x^2 - 1 - 3 = x^2 - 4$ ✓

Now we differentiate: $f'(x) = \dfrac{d}{dx}\left(x + 1 + \dfrac{-3}{x-1}\right) = 1 + \dfrac{3}{(x-1)^2}$.

Since $(x-1)^2 > 0$ for all $x \neq 1$, we have $\dfrac{3}{(x-1)^2} > 0$ for all $x$ in the domain. Therefore $f'(x) = 1 + \dfrac{3}{(x-1)^2} > 0$ for every $x \neq 1$.

The derivative is never zero and is undefined only at $x = 1$, which is not in the domain of $f$. Thus there are no critical numbers. Since $f'(x) > 0$ on both $(-\infty, 1)$ and $(1, \infty)$, the function is increasing on $(-\infty, 1)$ and increasing on $(1, \infty)$. Because the derivative never equals zero and never changes sign on the domain, there are no local extrema.

The answer is zero. Since $f'(x) = 1 + \dfrac{3}{(x-1)^2}$, it is always positive on the domain and never equals zero. Because there are no critical numbers and no sign change anywhere, the First Derivative Test guarantees no local maxima or minima exist.