극대와 극소

(a) Since $g'(x) = \frac{(x+4)(x-1)^2}{x-2}$, we see that $g'(x) = 0$ implies that $x = -4$ or $x = 1$. While $x = 2$ makes $g'$ undefined, we are told that $g$ has a vertical asymptote at $x = 2$, so $x = 2$ is not in the domain of $g$, and hence is technically not a critical number of $g$. Nonetheless, we place $x = 2$ on our first derivative sign chart since the vertical asymptote is a location at which $g'$ may change sign. (b) The first derivative sign chart shows that $g'(x) > 0$ for $x < -4$, $g'(x) < 0$ for $-4 < x < 1$, $g'(x) < 0$ for $1 < x < 2$, and $g'(x) > 0$ for $x > 2$. By the first derivative test, $g$ has a local maximum at $x = -4$ and neither a max nor min at $x = 1$. As these are the only two critical numbers, these are the only two locations for possible extremes. (Note: although $g$ changes from decreasing to increasing at $x = 2$, this is due to a vertical asymptote, and $g$ does not have a minimum there.) (c) Because $g$ is decreasing as $x \to 2^-$ (where $g$ has a vertical asymptote), $g$ does not have a global minimum. For $x > 2$, $g$ is always increasing, which suggests that $g$ does not have a global maximum (though we do not know for sure that $g$ increases without bound). (d) We observe that $$\begin{aligned}\lim_{x \to \infty} g'(x) =\mathstrut & \lim_{x \to \infty} \frac{(x+4)(x-1)^2}{x-2} \\ =\mathstrut & \lim_{x \to \infty} \frac{x^3 + 2x^2 - 7x + 4}{x-2} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} \\ =\mathstrut & \lim_{x \to \infty} \frac{x^2 + 2x - 7 + \frac{4}{x}}{1 - \frac{2}{x}} \\ =\mathstrut & \infty\end{aligned}$$ Since $g'(x) \to \infty$ as $x \to \infty$, this tells us that $g$ increases without bound as $x \to \infty$. (e) From all of our work above, we know that $g$ has a local maximum at $x = -4$, a horizontal tangent line with neither a max nor min at $x = 1$, and a vertical asymptote at $x = 2$, plus $g$ and $g'$ both increase without bound as $x \to \infty$. Thus, a possible graph of $g$ is the following.