Answer
$$
f(x)=x^{4}-4x^{3}
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) \neq f(x) \text { or }-f(x)
\end{aligned}
$$
No symmetry.
the critical numbers are: $x =0$, $x=3$.
Critical points:
$$ (0, 0) , \quad (3 ,-27)$$
Relative minimum at $3$
Neither a relative minimum nor maximum at 0 .
Also, we find that the function $f$ decreasing on $$(-\infty, 3)$$
and increasing on $$(3 , \infty) $$.
The points of inflection are
$$ ( 0 , 0 ) \quad \text{and} \quad ( 2 , -16)$$
$f(x)$ is concave upward on $(-\infty,0)$ and $(2, \infty)$.
$f(x)$ is concave downward on $(0 , 2)$ and
y-intercept:
$$
f(0)=0
$$
Work Step by Step
$$
f(x)=x^{4}-4x^{3}
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=(-x)^{4}-4(-x)^{3} \\
&=x^{4}+4 x^{3} \neq f(x) \text { or }-f(x)
\end{aligned}
$$
No symmetry.
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
&f^{\prime}(x)=4 x^{3}-12 x^{2}\\
&=4 x^{3}-12 x^{2}\\
&=4 x^{2}(x-3)
\end{aligned}
$$
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
$$
\begin{aligned}
&f^{\prime}(x)=4 x^{3}-12 x^{2}\\
&4 x^{3}-12 x^{2}=0\\
&4 x^{2}(x-3)=0\\
\Rightarrow \quad\quad\quad\quad\\
& x=0 , \quad\text{or}\quad x-3 =0 \\
& x=0 , \quad\text{or}\quad x =3 \\
\end{aligned}
$$
So,
the critical numbers are: $x =0$, $x=3$.
To find the critical points substituting in $f$ as follows:
$$
f(0)=(0)^{4}-4(0)^{3}=0
$$
$$
f(3)=(3)^{4}-4(3)^{3}=-27
$$
So, Critical points:
$$ (0, 0) , \quad (3 ,-27)$$
We can still apply the second derivative test, however, to find where $f$ is increasing and decreasing. as follows :
$$
\begin{array}{c}
f^{\prime \prime}(x)=12 x^{2}-24x \\
f^{\prime \prime}(3)=12 (3)^{2}-24(3) =36 \gt 0 \\
f^{\prime \prime}(0)=12 (0)^{2}-24(0) =0 \\
\end{array}
$$
we find that
Relative minimum at $3$
Second derivative test fails for 0,
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. We observe, the number $0$ divides the number line into two intervals: $(-\infty, 0)$ and $(0,3)$. use a test point in each of these intervals to find that sign of $f^{\prime} $ , as follows:
$$
\begin{array}{c}
f^{\prime}(-1)-4(-1)^{3}-12(-1)^{2}--16<0 \\
f^{\prime}(1)=4(1)^{3}-12(1)^{2}=-8<0
\end{array}
$$
So, we find that neither a relative minimum nor maximum at 0 .
Also, we find that the function $f$ decreasing on $$(-\infty, 3)$$
and increasing on $$(3 , \infty) $$.
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=12 x^{2}-24 x =0 \\
& 12 x(x-2) =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
&x =0 \text { or } x-2=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
&x =0 \text { or } x=2\\
\end{aligned}
$$
so $(0 , f(0)$ and $(2 , f(2)$ are inflection point.
Now ,we find $f(0) $ and f(2) as follows:
$$
f(0)=(0)^{4}-4(0)^{3}=0
$$
and
$$
f(2)=(2)^{4}-4(2)^{3}=-16
$$
So,the points of inflection are $$ ( 0 , 0 ) \quad \text{and} \quad ( 2 , -16)$$
$$
\begin{array}{l}
f^{\prime \prime}(-1)=12 (-1)^{2}-24 (-1)=36 \gt 0 \\
f^{\prime \prime}(1)=12 (1)^{2}-24 (1)=-12 \lt 0\\
f^{\prime \prime}(3)=12 (3)^{2}-24 (3)=36 \gt 0 \\
\end{array}
$$
Test a number in the intervals $(-\infty,0)$ and $(2, \infty)$ to find $f^{\prime\prime}(x)$ positive in these intervals, so $f(x)$ is concave upward on $(-\infty,0)$ and $(2, \infty)$.
Test a number in the interval $(0, 2)$ to find $f^{\prime\prime}(x)$ negative in that interval, so $f(x)$ is concave downward on $(0 , 2)$ and
y-intercept:
$$
f(0)=(0)^{4}-4(0)^{3}=0
$$
