Answer
The curve, observing from left to right,
- rises on $(-\infty,-\sqrt{3})$, concave down,
- reaches an absolute maximum at $(-\sqrt{3},5)$, turns
- at $(-1,1)$ changes concavity to "up" (inflection point),
- reaches a local minimum at $(0,-4)$, turns
- at $(1,1)$ changes concavity to "down" (inflection point),
- reaches an absolute maximum at $(\sqrt{3},5)$, turns
- and contiunes to fall on the interval $(\sqrt{3},\infty)$, concave down.
Work Step by Step
$y=f(x)=x^{2}(6-x^{2})-4$
$f'(x)=-4x^{3}+12x=-4x(x^{2}-3)=-4x(x+\sqrt{3})(x-\sqrt{3})$
$f''(x)=-12x^{2}+12=-12(x^{2}-1)=-12(x+1)(x-1)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=0, \pm\sqrt{3}\qquad$... (critical points)
$\left[\begin{array}{ccccccc}
y': & ++ & | & -- & | & ++ & | & --\\
& & -\sqrt{3} & & 0 & & \sqrt{3} & \\
y: & \nearrow & \max & \searrow & \min & \nearrow & \max & \searrow
\end{array}\right] \left\{\begin{array}{l}
f(-\sqrt{3})=5\\
f(0)=-4\\
f(\sqrt{3})=5
\end{array}\right.$
Analyzing $f''(x)$,
$f''(x)=0$ for $ x=\pm 1,\qquad$
$\left\{\begin{array}{ll}
f(-1)=1, & (-1,1)\\
f(1)=1, & (-1,1)
\end{array}\right. \quad$ ... points of inflection.
Concavity and points of inflection:
$\left[\begin{array}{cccccc}
y'''': & -- & | & ++ & -1 & -- & & \\
& & -1 & & 1 & & & \\
y: & \cap & i.p. & \cup & i.p. & \cap & &
\end{array}\right]\left\{\begin{array}{l}
f(-1)=1\\
\\
f(1)=1
\end{array}\right.$
