§3.3 Using derivatives to identify extreme values (continued)

Summary

A graph of $f'$ shows $f'$ negative for $x \lt -1$, positive for $-1 \lt x \lt 1$, and positive for $x \gt 1$, with $f(0) = -\frac{1}{2}$. The function $f$ has

The same graph of $f'$ shows horizontal tangents at about $x = -0.35$ and $x = 1$, with $f'$ steepest between them. The inflection points of $f$ occur at

Suppose $g'(2) = 0$ and $g'(x) \gt 0$ on both $1 \lt x \lt 2$ and $2 \lt x \lt 3$. At $x = 2$ the function $g$ has

With $g'(2) = 0$, $g' \gt 0$ nearby, and $g''$ existing on $1 \lt x \lt 3$, the value $x = 2$ is special because

A graph of $h'$ shows $h' \lt 0$ for $x \lt 0$ and $h' \gt 0$ for $x \gt 0$, with $h'$ tending to $3$ as $x \to \infty$. If $h(x) = 0$ has two distinct real solutions, their signs must be

With the same $h'$ approaching $3$ as $x \to \infty$, the graph of $y = h(x)$ for large $x$ looks like

Let $g'(x) = \frac{(x+4)(x-1)^2}{x-2}$ with $g$ having a vertical asymptote at $x = 2$. The critical numbers of $g$ and their classification are

For the same $g$, the value of $\lim_{x \to \infty} g'(x)$ and its meaning are

Let $p''(x) = (x+1)(x-2)e^{-x}$. The critical number $x = \frac{\sqrt{5}-1}{2} \approx 0.618$ lies between $-1$ and $2$, so $p$ has

With the same $p''(x) = (x+1)(x-2)e^{-x}$, the inflection points of $p$ occur at

The point $(2, \frac{12}{e^2})$ lies on $y = p(x)$ with $p'(2) = -\frac{5}{e^2}$, and $x = 2$ is an inflection point. The tangent line at $x = 2$ lies

For $p''(x) = (x+1)(x-2)e^{-x}$, compute $p''(0)$.

The critical number $\frac{\sqrt{5}-1}{2}$ appears in the study of $p$. Compute its decimal value, rounded to three places.

If $g''(2) = 0$ and $g''$ changes sign at $x = 2$, then $g$ has an inflection point at $x = 2$.