극대와 극소

(a) Note that $p''(x)=0$ at $x=-1$ and $x=2$. The second derivative sign chart for $p$ shows that $p''(x)$ is negative for $-1 < x < 2$ and positive for all other values of $x$. Since $p''$ changes sign at $x = -1$ and $x = 2$, $p$ has points of inflection at $x = -1$ and $x = 2$. In particular, $p$ is concave down for $-1 < x < 2$ and concave up for all other values of $x$. (b) If we compute $p''\left(\frac{\sqrt{5}-1}{2} \right)$, we find that $p''\left(\frac{\sqrt{5}-1}{2} \right) \approx -1.205$, which means that $p$ is concave down at $x = \frac{\sqrt{5}-1}{2}$. Since $p$ has a horizontal tangent line at this point, it follows that $p$ has a local maximum at $x = \frac{\sqrt{5}-1}{2}$ by the second derivative test. (c) Since $p(2) = \frac{12}{e^2}$ and $p'(2) = -\frac{5}{e^2}$, the equation of the tangent line to $p$ at $x = 2$ is given by $y = -\frac{5}{e^2}(x-2) + \frac{12}{e^2}$. To determine whether the tangent line lies above or below the graph of $y=p(x)$ at $x = 2$, we consider the concavity of $p$. Recall we know that $p''(2) = 0$ and that $p$ is concave down to the left of $x = 2$ and concave up to the right. Thus, the tangent line lies neither solely below nor solely above the graph of $y = p(x)$ at $x = 2$; it lies below $y = p(x)$ to the right of $x = 2$ and above $y = p(x)$ to the left of $x = 2$.