Answer
$$
f(x)=x^{5}-15x^{3}
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=(-x)^{5}-15(-x)^{3}=-f(x)
\end{aligned}
$$
The graph is symmetric about the origin.
The critical numbers are: $x =0$, $x=-3$, and $x=3$ .
Critical point:
$$
(0,0) , \quad (-3 ,162), \quad \text {and} \quad (3 , -162)
$$
No relative extreme at $0$ , relative minimum at $3,$ and relative maximum at $ -3$
The function $f$ Increasing on
$$
(-\infty, -3) \quad \text{and} \quad(3,\infty)
$$
Decreasing on $$(-3, 3)$$
$(0 , f(0))$ and $(\frac{3}{\sqrt 2} , f(\frac{3}{\sqrt 2}))$, and $(\frac{-3}{\sqrt 2} , f(\frac{-3}{\sqrt 2}))$ are inflection point.
The points of inflection are $$ ( 0 , 0 ) \quad, (\frac{3}{\sqrt 2}, -100.23) \quad \text{and} \quad (\frac{-3}{\sqrt 2}, 100.23)$$
$f(x)$ is concave downward on $(-\infty,\frac{-3}{\sqrt 2})$ and $(0,\frac{3}{\sqrt 2} )$.
$f(x)$ is concave upward on $(\frac{-3}{\sqrt 2}, 0 )$ and $(\frac{3}{\sqrt 2} , \infty )$ .
y-intercept:
$$
f(0)=(0)^{5}-15(0)^{3}=0
$$
$x$ -intercepts:
$$
\begin{aligned}
f^{\prime\prime}(x)&= x^{5}-15 x^{3}=0\\
&= x^{3} (x^{2}-15 )=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x^{3}=0 \quad \text{or} \quad (x^{2}-15 )=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0, \quad x=\pm \sqrt{15} \\
\end{aligned}
$$
Work Step by Step
$$
f(x)=x^{5}-15x^{3}
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=(-x)^{5}-15(-x)^{3}\\
&=-x^{5}+15 x^{3}\\
&=-f(x)
\end{aligned}
$$
The graph is symmetric about the origin.
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x)&= 5 x^{4}-45 x^{2} \\
&= 5 x^{2}\left(x^{2}-9\right) \\
&=5 x^{2}(x+3)(x-3)
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) & =5 x^{2}(x+3)(x-3)=0 \\
\Rightarrow \quad\quad\quad\quad\\
& x=0 , \quad\text{or}\quad x+3=0 ,\quad \text{or} \quad x-3 =0 \\
& x=0 , \quad\text{or}\quad x=-3 ,\quad \text{or} \quad x=3
\end{aligned}
$$
So,
the critical numbers are: $x =0$, $x=-3$, and $x=3$ .
To find the critical points substituting in $f$ as follows:
$$
f(0)=(0)^{5}-15(0)^{3}=0
$$
$$
f(-3)=(-3)^{5}-15(-3)^{3}=162
$$
$$
f(3)=(3)^{5}-15(3)^{3}=-162
$$
So, Critical point:
$$
(0,0) , \quad (-3 ,162), \quad \text {and} \quad (3 , -162)
$$
We can still apply the second derivative test, however, to find where $f$ is increasing and decreasing. as follows :
$$
\begin{array}{c}
f^{\prime \prime}(x)=20 x^{3}-90 x \\
f^{\prime \prime}(- 3)=20 (-3)^{3}-90 (-3) =-270 \lt 0 \\
f^{\prime \prime}(0)=20 (0)^{3}-90 (0)= 0 \\
f^{\prime \prime}(3)=20 (3)^{3}-90 (3) =270 \gt 0
\end{array}
$$
we find that
No relative extreme at $0$ , relative minimum at $3,$ and relative maximum at $ -3$
So, the function $f$ Increasing on
$$
(-\infty, -3) \quad \text{and} \quad(3,\infty)
$$
Decreasing on $$(-3, 3)$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=20 x^{3}-90 x=0 \\
&=10 x\left(2x^{2}-9\right)=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x\left(2x^{2}-9\right) =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0 , \quad\text{or}\quad x=\frac{-3}{\sqrt 2},\quad \text{or} \quad x=\frac{3}{\sqrt 2}
\end{aligned}
$$
so $(0 , f(0))$ and $(\frac{3}{\sqrt 2} , f(\frac{3}{\sqrt 2}))$, and $(\frac{-3}{\sqrt 2} , f(\frac{-3}{\sqrt 2}))$ are inflection point.
Now ,we find $f(0) $, f(\frac{3}{\sqrt 2}) and f(\frac{-3}{\sqrt 2}) as follows:
$$
f(0)=(0)^{5}-15(0)^{3}=0
$$
,
$$
f(\frac{3}{\sqrt 2})=(\frac{3}{\sqrt 2})^{5}-15(\frac{3}{\sqrt 2})^{3}=(\frac{3}{\sqrt 2})^{5}-15(\frac{3}{\sqrt 2})^{3} \approx -100.23
$$
and
$$
f(\frac{-3}{\sqrt 2})=(\frac{-3}{\sqrt 2})^{5}-15(\frac{-3}{\sqrt 2})^{3}=(\frac{-3}{\sqrt 2})^{5}-15(\frac{-3}{\sqrt 2})^{3} \approx 100.23
$$
So,the points of inflection are $$ ( 0 , 0 ) \quad, (\frac{3}{\sqrt 2}, -100.23) \quad \text{and} \quad (\frac{-3}{\sqrt 2}, 100.23)$$
$$
\begin{array}{l}
f^{\prime \prime}(-3)=20 (-3)^{3}-90 (-3)=-270 \lt 0 \\
f^{\prime \prime}(-2)=20 (-2)^{3}-90 (-2)=20 \gt 0\\
f^{\prime \prime}(3)=-12 (-2)^{2}+12=270 \gt 0 \\
f^{\prime \prime}(2)=-12 (-2)^{-2}+12=-20 \lt 0 \\
\end{array}
$$
Test a number in the intervals $(-\infty,\frac{-3}{\sqrt 2})$ and $(0,\frac{3}{\sqrt 2} )$ we find $f^{\prime\prime}(x)$ is negative in these intervals, so $f(x)$ is concave downward on $(-\infty,\frac{-3}{\sqrt 2})$ and $(0,\frac{3}{\sqrt 2} )$.
Test a number in the interval $(\frac{-3}{\sqrt 2}, 0 )$ and $(\frac{3}{\sqrt 2} , \infty )$ to find $f^{\prime\prime}(x)$ positive in these intervals, so $f(x)$ is concave upward on $(\frac{-3}{\sqrt 2}, 0 )$ and $(\frac{3}{\sqrt 2} , \infty )$ .
y-intercept:
$$
f(0)=(0)^{5}-15(0)^{3}=0
$$
$x$ -intercepts:
$$
\begin{aligned}
f^{\prime\prime}(x)&= x^{5}-15 x^{3}=0\\
&= x^{3} (x^{2}-15 )=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x^{3}=0 \quad \text{or} \quad (x^{2}-15 )=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0, \quad x=\pm \sqrt{15} \\
\end{aligned}
$$
