Answer
$$
f(x)=16x+\frac{1}{x^{2}}=16x+x^{-2}
$$
The domain is
$$(-\infty, 0) \cup(0, \infty)$$
The graph is not symmetric about the y-axis or the origin.
the critical numbers are: $x =\frac{1}{2} $ .
Critical point:
$$
(\frac{1}{2} , 12) .
$$
$f^{\prime}(x) $ does not exist at $x=0$
So, the function $f$ Increasing on
$$
(-\infty, 0) \quad \text{and} \quad( \frac{1}{2} ,\infty)
$$
Decreasing on $$(0, \frac{1}{2} ).$$
and we find that:
relative minimum at $ \frac{1}{2},$
There are no inflection points.
$f(x)$ is concave upward on $(-\infty, 0)$ and $(0, \infty )$ .
there are no x-intercepts.
there is no y-intercept.
Vertical asymptotic at $x=0$
$$
\lim\limits_{x \to 0}f(x)=\infty.
$$
Work Step by Step
$$
f(x)=16x+\frac{1}{x^{2}}=16x+x^{-2}
$$
since $f(x)$ does not exist when $x=0,$ the domain is
$$(-\infty, 0) \cup(0, \infty)$$
$$
\begin{aligned}
f(-x) &=16(-x)+\frac{1}{(-x)^{2}}\\
&=-16x+x^{-2}\\
\end{aligned}
$$
The graph is not symmetric about the y-axis or
the origin.
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x)&= 16 -2 x^{-3}\\
&=16-\frac{2}{x^{3}} \\
&=\frac{2\left(8 x^{3}-1\right)}{x^{3}}
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) & =\frac{2\left(8 x^{3}-1\right)}{x^{3}} =0 \\
\Rightarrow \quad\quad\quad\quad\\
\left(8 x^{3}-1\right)\\
\Rightarrow \quad\quad\quad\quad\\
x=\frac{1}{2}
\end{aligned}
$$
So,
the critical numbers are: $x =\frac{1}{2} $ .
To find the critical points substituting in $f$ as follows:
$$
f(\frac{1}{2})=16(\frac{1}{2})+\frac{1}{(\frac{1}{2})^{2}}=12
$$
So, Critical point:
$$
(\frac{1}{2} , 12) .
$$
$f^{\prime}(x) $ does not exist at $x=0$
Now, we can use the first derivative test. Check the sign of $f^{\prime}(x)$. in the intervals
$$
(-\infty, 0), \quad ( 0, \frac{1}{2} ), \quad \text {and } ( \frac{1}{2} , \infty) .
$$
(1)
Test a number in the interval $(-\infty, 0) $ say $-1$:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{2\left(8 (-1)^{3}-1\right)}{(-1)^{3}} \\
&=18 \\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, 0) $.
(2)
Test a number in the interval $( 0, \frac{1}{2} ) $ say $\frac{1}{3}$:
$$
\begin{aligned}
f^{\prime}(\frac{1}{3}) &=\frac{2\left(8 (\frac{1}{3})^{3}-1\right)}{(\frac{1}{3})^{3}}\\
&=-38 \\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( 0 , \frac{1}{2} ) $.
(3)
Test a number in the interval $( \frac{1}{2}, \infty) $ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{2\left(8 (1)^{3}-1\right)}{(1)^{3}}\\
&=14 \\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $( \frac{1}{2} , \infty ) $.
So, the function $f$ Increasing on
$$
(-\infty, 0) \quad \text{and} \quad( \frac{1}{2} ,\infty)
$$
Decreasing on $$(0, \frac{1}{2} ).$$
and we find that:
relative minimum at $ \frac{1}{2},$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=6x^{-4} \\
\Rightarrow\quad\quad\quad\quad\quad\\
\text { is never equal to zero.}
\end{aligned}
$$
so
There are no inflection points.
(Recall that $f(x)$ does not exist at $x=0$)
Now, we can use the second derivative test. Check the sign of $f^{\prime\prime}(x)$. in the intervals
$$
(-\infty, 0), \quad \quad ( 0 ,\infty ).
$$
$$
\begin{array}{l}
f^{\prime \prime}(-1)=6(-1)^{-4} =6 \gt 0 \\
f^{\prime \prime}(1)=6(1)^{-4} =6 \gt 0 \\
\end{array}
$$
Test a number in the interval $(-\infty, 0)$ and $(0, \infty )$ we find $f^{\prime\prime}(x)$ is positive in these intervals, so $f(x)$ is concave upward on $(-\infty, 0)$ and $(0, \infty )$ .
Since $f(x)$ is never zero, so there are no x-intercepts.
y-intercept:
Since $f(x)$ does not exist for $x=0 $ so there is no y-intercept.
Vertical asymptotic at $x=0$
$$
\lim\limits_{x \to 0}f(x)=\infty
$$
