Answer
$$
f(x)=-x^{4}+6x^{2}
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) =f(x)
\end{aligned}
$$
symmetry about the y-axis.
The critical numbers are: $x =0$, $x=\sqrt{3}$, and $x=-\sqrt{3}$ .
Critical point:
$$ (0, 0) , \quad (-\sqrt{3} ,9), \quad \text {and} \quad (\sqrt{3} , 9)$$
Relative minimum at $0,$ relative maximum at $ \sqrt{3}$ and $- \sqrt{3}$
The function $f$ Increasing on
$$(-\infty, -\sqrt {3}) \quad \text{and} \quad(0, \sqrt {3})$$
Decreasing on
$$(-\sqrt {3},0) \quad \text{and} \quad (\sqrt {3}, \infty )$$
The points of inflection are $$ ( 1 , 5 ) \quad \text{and} \quad ( -1 , 5 )$$
$f(x)$ is concave downward on $(-\infty,-2)$ and $(2, \infty)$.
$f(x)$ is concave upward on $(-1,1)$ and
y-intercept:
$$
f(0)=0
$$
Work Step by Step
$$
f(x)=-x^{4}+6x^{2}
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=-(-x)^{4}+6(-x)^{2}\\
&=-x^{4}+6 x^{2}\\
&=f(x)
\end{aligned}
$$
symmetry about the y-axis.
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) & =4 x^{3}+12 x \\
&-4 x^{3}+12 x=0 \\
&-4 x\left(x^{2}-3\right)=0 \\
&-4 x(x-\sqrt{3})(x+\sqrt{3}) =0
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) & =-4 x(x-\sqrt{3})(x+\sqrt{3}) =0 \\
\Rightarrow \quad\quad\quad\quad\\
& x=0 , \quad\text{or}\quad x-\sqrt{3}=0 ,\quad \text{or} \quad x+\sqrt{3}
=0 \\
& x=0 , \quad\text{or}\quad x=\sqrt{3} ,\quad \text{or} \quad x=-\sqrt{3}
\end{aligned}
$$
So,
the critical numbers are: $x =0$, $x=\sqrt{3}$, and $x=-\sqrt{3}$ .
To find the critical points substituting in $f$ as follows:
$$
f(0)=-(0)^{4}+6(0)^{2}=0
$$
$$
f(\sqrt{3})=-(\sqrt{3})^{4}+6(\sqrt{3})^{2}=9
$$
$$
f(-\sqrt{3})=-(-\sqrt{3})^{4}+6(-\sqrt{3})^{2}=9
$$
So, Critical point:
$$ (0, 0) , \quad (-\sqrt{3} ,9), \quad \text {and} \quad (\sqrt{3} , 9)$$
We can still apply the second derivative test, however, to find where $f$ is increasing and decreasing. as follows :
$$
\begin{array}{c}
f^{\prime \prime}(x)=-12 x^{2}+12 \\
f^{\prime \prime}(- \sqrt{3})=-12 (- \sqrt{3})^{2}+12=-24\lt 0 \\
f^{\prime \prime}(0)=-12 (0)^{2}+12=12 \gt 0 \\
f^{\prime \prime}(- \sqrt{3})=-12 (- \sqrt{3})^{2}+12=-24\lt 0 \\
\end{array}
$$
we find that
Relative minimum at $0,$ relative maximum at $ \sqrt{3}$ and $- \sqrt{3}$
So, the function $f$ Increasing on
$$(-\infty, -\sqrt {3}) \quad \text{and} \quad(0, \sqrt {3})$$
Decreasing on
$$(-\sqrt {3},0) \quad \text{and} \quad (\sqrt {3}, \infty )$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=-12 x^{2}+12 =0 \\
&-12\left(x^{2}-1\right) =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x^{2}-1 =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=\pm 1
\end{aligned}
$$
so $(1 , f(1)$ and $(-1 , f(-1)$ are inflection point.
Now ,we find $f(1) $ and f(-1) as follows:
$$
f(1)=-(1)^{4}+6(1)^{2}=5
$$
and
$$
f(-1)=-(-1)^{4}+6(-1)^{2}=5
$$
So,the points of inflection are $$ ( 1 , 5 ) \quad \text{and} \quad ( -1 , 5 )$$
$$
\begin{array}{l}
f^{\prime \prime}(-2)=-12 (-2)^{2}+12=-36 \lt 0 \\
f^{\prime \prime}(0)=-12 (0)^{2}+12=12 \gt 0\\
f^{\prime \prime}(2)=-12 (-2)^{-2}+12=-36 \lt 0 \\
\end{array}
$$
Test a number in the intervals $(-\infty,-1)$ and $(1, \infty)$ to find $f^{\prime\prime}(x)$ negative in these intervals, so $f(x)$ is concave downward on $(-\infty,-2)$ and $(2, \infty)$.
Test a number in the interval $(-1,1)$ to find $f^{\prime\prime}(x)$ positive in that interval, so $f(x)$ is concave upward on $(-1,1)$ and
y-intercept:
$$
f(0)=-(0)^{4}+6(0)^{2}=0
$$
