Answer
$$
f(x)=x^{3} - 6 x^{2} + 12 x- 11
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) \ne f(x)
\end{aligned}
$$
No symmetry
The only critical number is: $x =2$.
Critical point:
$$ (2, -3) . $$
The function $f$ is increasing on interval $ ( -\infty , \infty) . $
No relative extremum at $ (2, -3) . $
$(2 , -3)$ is inflection point.
$f(x)$ is concave downward on $(-\infty, 2 )$.
$f(x)$ is concave upward on $(2, \infty )$.
y-intercept:
$$
f(0)=-11
$$
Work Step by Step
$$
f(x)=x^{3} - 6 x^{2} + 12 x- 11
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=(-x)^{3}-6(-x)^{2}+12(-x)-11 \\
&=-x^{3}-6 x^{2}-12 x-11 \\
& \ne f(x)
\end{aligned}
$$
No symmetry
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) & =3 x^{2}-12 x+12 \\
&=3\left(x^{2}-4 x+4\right)\\
&=3(x-2)^{2}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) & =3(x-2)^{2}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x-2=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
&x=2 \\
\end{aligned}
$$
So,
the only critical number is: $x =2$.
To find the critical points substituting in $f$ as follows:
$$
f(2)=(2)^{3} - 6 (2)^{2} + 12 (2)- 11=-3
$$
So, Critical point:
$$ (2, -3) . $$
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. Now, check the sign of $f^{\prime}(x)$. in the intervals
$$
(-\infty,2 ), \quad\quad (2 , \infty ).
$$
(1)
Test a number in the interval $(-\infty, 2 )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=3((0)-2)^{2} \\
&=12 \\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, 2 )$.
(2)
Test a number in the interval $(2, \infty )$ say $3$:
$$
\begin{aligned}
f^{\prime}(3) &=3((3)-2)^{2} \\
&=3 \\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(2, \infty )$.
So, the function $f$ is increasing on interval $ ( -\infty , \infty) . $
No relative extremum at $ (2, -3) . $
Now, the second derivative is
$$
\begin{aligned}
f^{\prime \prime}(x) &=6 x-12 \\
&=6( x-2) \\
\end{aligned}
$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=6( x-2)=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x-2 =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=2
\end{aligned}
$$
so $(2 , -3)$ is inflection point.
*
$$
\begin{array}{l}
f^{\prime \prime}(0)=-6( (0)-2)=-12\lt 0 \\
f^{\prime \prime}(3)=6( (3)-2)=6\gt 0
\end{array}
$$
Test a number in the interval $(-\infty, 2 )$ to see that $f^{\prime\prime}(x)$ is negative in that interval, so $f(x)$ is concave downward on $(-\infty, 2 )$.
**
Test a number in the interval $(2, \infty )$to find $f^{\prime\prime}(x)$ positive in that interval, so $f(x)$ is concave upward on $(2, \infty )$.
y-intercept:
$$
f(0)=(0)^{3} - 6 (0)^{2} + 12 (0)- 11=-11
$$
