Answer
The curve, observing from left to right,
- rises on $(-\infty,2)$, is concave down,
- rising, crosses the y-xis at $(0,-7)$,
- still rising, crosses the x-axis at x=1,
- at $(2,1)$ changes concavity to "up" (inflection point),
- and contiunes to rise on the interval $(2,\infty)$, concave up.
No extrema; inflection point at $(2,1)$.
Work Step by Step
$y=f(x)=(x-2)^{3}+1$
$f'(x)=3( x-2)^{2}$
$f''(x)=6( x-2)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=2\qquad$... (critical point)
$f'$ is nonnegative, so $f$ is always rising.
$f$ has no extrema
$f''(x)=0$ for $x=2,\qquad f(2)=1,\quad(2,1)$ is a point of inflection.
For graphing purposes:
$f(0)=-7,\qquad f(1)=0$
Concavity and points of inflection:
$\left[\begin{array}{lllllll}
f''(x): & -\infty & -- & 2 & ++ & \infty & \\
f(x): & & \cap & inf. & \cup & &
\end{array}\right]$
The curve, observing from left to right,
- rises on $(-\infty,2)$, is concave down,
- rising, crosses the y-xis at $(0,-7)$,
- still rising, crosses the x-axis at x=1,
- at $(2,1)$ changes concavity to "up" (inflection point),
- and contiunes to rise on the interval $(2,\infty)$, concave up.
No extrema; inflection point at $(2,1)$.
