Answer
$$
f(x)=x^{4}-24 x^{2}+80
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) =f(x)
\end{aligned}
$$
symmetry about the y-axis.
The critical numbers are: $x =0$, $x=2 \sqrt{3}$, and $x=-2 \sqrt{3}$ .
Critical point:
$$ (0,80) , \quad (2 \sqrt{3}, -64), \quad \text {and} \quad (-2 \sqrt{3}, -64)$$
Relative maximum at $0,$ relative minimum at $-2 \sqrt{3}$ and $2 \sqrt{3}$
The function $f$ Increasing on
$$(-2 \sqrt{3}, 0) \quad \text{and} \quad(2 \sqrt{3}, \infty)$$
Decreasing on
$$(-\infty,-2 \sqrt{3}) \quad \text{and} \quad (0,2 \sqrt{3})$$
Points of inflection are $$ ( 2 , 0 ) \quad \text{and} \quad ( -2 , 0 )$$
$f(x)$ is concave upward on $(-\infty,-2)$ and $(2, \infty)$.
$f(x)$ is concave downward on $(-2,2)$ .
y-intercept:
$$
f(0)=80
$$
$x$ -intercepts:
$$
x=\pm 2 , \quad\text{or}\quad x=\pm 2 \sqrt{5}
$$
Work Step by Step
$$
f(x)=x^{4}-24 x^{2}+80
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=(-x)^{4}-24(-x)^{2}+80 \\
&=x^{4}-24 x^{2}+80\\
&=f(x)
\end{aligned}
$$
symmetry about the y-axis.
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) & =4 x^{3}-48 x \\
&= 4 x\left(x^{2}-12\right)\\
& = 4 x(x-2 \sqrt{3})(x+2 \sqrt{3}) \\
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) & =4 x(x-2 \sqrt{3})(x+2 \sqrt{3}) =0 \\
\Rightarrow \quad\quad\quad\quad\\
& x=0 , \quad\text{or}\quad x-2 \sqrt{3}=0 ,\quad \text{or} \quad x+2 \sqrt{3}=0 \\
& x=0 , \quad\text{or}\quad x=2 \sqrt{3} ,\quad \text{or} \quad x=-2 \sqrt{3} \\
\end{aligned}
$$
So,
the critical numbers are: $x =0$, $x=2 \sqrt{3}$, and $x=-2 \sqrt{3}$ .
To find the critical points substituting in $f$ as follows:
$$
f(0)=(0)^{4}-24 (0)^{2}+80=80
$$
$$
f(2 \sqrt{3})=(2 \sqrt{3})^{4}-24 (2 \sqrt{3})^{2}+80=-64
$$
$$
f(-2 \sqrt{3})=(-2 \sqrt{3})^{4}-24 (-2 \sqrt{3})^{2}+80=-64
$$
So, Critical point:
$$ (0,80) , \quad (2 \sqrt{3}, -64), \quad \text {and} \quad (-2 \sqrt{3}, -64)$$
We can still apply the second derivative test, however, to find where $f$ is increasing and decreasing. as follows :
$$
\begin{array}{c}
f^{\prime \prime}(x)=12 x^{2}-48 \\
f^{\prime \prime}(-2 \sqrt{3})=12(-2 \sqrt{3})^{2}-48=96>0 \\
f^{\prime \prime}(0)=12(0)^{2}-48=-48<0 \\
f^{\prime \prime}(2 \sqrt{3})=12(2 \sqrt{3})^{2}-48=96>0
\end{array}
$$
we find that
Relative maximum at $0,$ relative minimum at $-2 \sqrt{3}$ and $2 \sqrt{3}$
So, the function $f$ Increasing on
$$(-2 \sqrt{3}, 0) \quad \text{and} \quad(2 \sqrt{3}, \infty)$$
Decreasing on
$$(-\infty,-2 \sqrt{3}) \quad \text{and} \quad (0,2 \sqrt{3})$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=12x^{2}-48=0 \\
&=12(x^{2}-4)=0
\Rightarrow\quad\quad\quad\quad\quad\\
& x^{2}-4 =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=2 , \quad\text{or}\quad x=-2
\end{aligned}
$$
so $(2 , f(2)$ and $(-2 , f(-2)$ are inflection point.
Now ,we find $f(2) $ and f(-2) as follows:
$$
f(2)=(2)^{4}-24 (2)^{2}+80=0
$$
and
$$
f(-2)=(-2)^{4}-24 (-2)^{2}+80=0
$$
So,the points of inflection are $$ ( 2 , 0 ) \quad \text{and} \quad ( -2 , 0 )$$
$$
\begin{array}{l}
f^{\prime \prime}(-3)=12 (-3)^{2}-48=60 \gt 0 \\
f^{\prime \prime}(-1)=12 (-1)^{2}-48-36 \lt 0\\
f^{\prime \prime}(1)=12 x^{2}-48 =-36 \lt 0 \\
f^{\prime \prime}(3)=12 x^{2}-48 =60 \gt 0 \\
\end{array}
$$
Test a number in the intervals $(-\infty,-2)$ and $(2, \infty)$ to find $f^{\prime\prime}(x)$ positive in these intervals, so $f(x)$ is concave upward on $(-\infty,-2)$ and $(2, \infty)$.
Test a number in the interval $(-2,2)$ to find $f^{\prime\prime}(x)$ negative in that interval, so $f(x)$ is concave downward on $(-2,2)$ and
y-intercept:
$$
f(0)=(0)^{4}-24 (0)^{2}+80=80
$$
$x$ -intercepts:
$$
\begin{aligned}
0&=x^{4}-24 x^{2}+80 \\
\text{Let} \ \ \ \ u=x^{2} \ \ \ \ \text{then}\\
& u^{2}-24 u+80=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& (u-4)(u-20)=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& u=4 , \quad\text{or}\quad u=20 \\
&x=\pm 2 , \quad\text{or}\quad x=\pm 2 \sqrt{5}
\end{aligned}
$$
