Answer
The curve, observing from left to right,
- falls on $(-\infty,-1)$, is concave up,
- at $(-1,1)$ changes concavity to "down" (inflection point),
- and contiunes to fall on the interval $(-1,\infty)$, concave down.
- (crosses the the x-axis at the origin)
No extrema; inflection point at $(-1,1)$.
Work Step by Step
$y=f(x)=1-(x+1)^{3}$
$f'(x)=-3( x+1)^{2}$
$f''(x)=-6( x+1)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=2\qquad$... (critical point)
$f'$ is always negative or 0, so $f$ is always falling
$f$ has no extrema
$f''(x)=0$ for $x=-1,\qquad f(-1)=1,\quad(-1,1)$ is a point of inflection.
For graphing purposes:
$ f(0)=0,\qquad$
Concavity and points of inflection:
$\left[\begin{array}{cccccc}
f''(x): & -\infty & ++ & -1 & -- & \infty & \\
f(x): & & \cup & inf. & \cap & &
\end{array}\right]$
