Answer
The graph, observing from left to right,
- falls on $(-\infty,1)$, is concave up,
- at $(0,-3)$ reaches a local minimum, turns down,
- starts increasing on the interval $(0,2),$ still concave up,
- at $(1,1)$ changes concavity to "down" (inflection point), and continues rising to
- the point $(2,5)$, a local maximum, from where
- it falls on the interval $(3,\infty)$, concave down.
Work Step by Step
$y=f(x)=-2x^{3}+6x^{2}-3$
$f'(x)=-6x^{2}+12x=-6x(x-2)$
$f''(x)=-12x+12=-12(x-1)$
Analyzing $f'(x)$, we see that it is defined everywhere, and has zeros $ x=0,2\qquad$... (critical points)
Applying the 2nd derivative test,
$\left\{\begin{array}{lllll}
x=0, & f''(0)=12\gt 0 & \Rightarrow(\min), & f(0)=-3 & (0-3)\\
x=2, & f''(0)=-12 \lt 0 & \Rightarrow(\max), & f(2)=5 & (2,5)
\end{array}\right.$
$f''(x)=0$ for $x=1,\qquad f(1)=1,\quad(1,1)$ is a point of inflection.
For graphing purposes:
The graph of $f'(x)$ is a parabola that opens down. The graph is above the x-axis between the zeros, and below for the other two intervals.
$\left[\begin{array}{ccccccccccc}
f'(x): & -\infty & -- & 0 & ++ & 2 & -- & \infty\\
f(x): & & \searrow & | & \nearrow & | & \searrow &
\end{array}\right]$
Concavity and points of inflection:
$\left[\begin{array}{cccccccc}
f''(x): & -\infty & ++ & 1 & -- & \infty & \\
f(x): & & \cup & inf. & \cap & &
\end{array}\right]$
