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We are given the functions
$$
f(x)=\frac{12x^2-16}{x^3},\qquad f'(x)=-\frac{12(x^2-4)}{x^4},\qquad f''(x)=\frac{24(x^2-8)}{x^5}.
$$
The questions below are about the function $f(x)$. Answer parts (1) through (10) below. If the requested feature is missing, then explain why. Be sure to include the work/test that you used to rigorously reach your conclusion. It is not sufficient to refer to the graph.
(1) State the function's domain.
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(1) Since, for $x=0$ the given function is undefined. The domain of $f(x)$ is (-∞,0) U (0,∞).

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(2) Find all $x$- and $y$-intercepts.
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(2) We have to plug $f(x)=0$
$\frac{12x^2-16}{x^3}=0$
$x^2=\frac{16}{12}=\frac{4}{3}$
$x=\frac{2}{\sqrt3}$ take the radical out of the denominator
$x=\frac{2\sqrt3}{3}$
The x-intercepts are $(\frac{2\sqrt3}{3}, 0)$ and $(\frac{-2\sqrt3}{3}, 0)$

For the y-intercepts we have to plug $x=0$ but $f(x)$ is not defined at $x=0$ hence, y-intercepts do not exist.

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(3) Find all equations of horizontal asymptotes.
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(3)Line $y=L$ is a horizontal asymptote of the function $y=f(x)$, if either $\lim_{h\to∞} f(x)=L$ or $\lim_{h\to-∞} f(x)=L$ and $L$ is finite.
$\lim_{h\to∞}(\frac{12x^2-16}{x^3})$
$\lim_{h\to∞}(\frac{12}{x}-\frac{16}{x^3})=0$
and
$\lim_{h\to-∞}(\frac{12x^2-16}{x^3})$
$\lim_{h\to-∞}(\frac{12}{x}-\frac{16}{x^3})=0$
hence, the horizontal asymptote is $y=0$

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(4) Find all equations of vertical asymptotes.
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(4) The line $x=L$ is a vertical asymptote of the function $y=f(x)$ if the limit of the function (one-sided) at this point is infinite.
Check at $x=0$
$\lim_{h\to0}(\frac{12x^2-16}{x^3})$
$\lim_{h\to0}(\frac{12}{x}-\frac{16}{x^3})=-∞$
Since the limit is infinite then $x=0$ is a vertical asymptote

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(5) Find the interval(s) where $f$ is increasing.
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(5)The derivative of the function will allow us to see where $f(x)$ is increasing. The derivative is
$f'(x)=-\frac{12(x^2-4)}{x^4}$
This can be factored out to
$f(x)=-\frac{12(x+2)(x-2)}{x^4}$
Set the equations to equal to zero
$x+2=0$
$x-2=0$
$-12=0$
The critical points of the function are $x=0,2-2$
The critical points are placed on a number line and integars above the critical points are plugged into the first derivative of the function to find the sign (positive or negative) of the slope.
The numbers I choose to plug into the first derivative are -3,-1,1,and 3.
The first derivative is
$f'(x)=-\frac{12(x^2-4)}{x^4}$
$f'(-3)=-\frac{12(-3^2-4)}{-3^4}=-$
$f'(-1)=-\frac{12(-1^2-4)}{-1^4}=-$
$f'(1)=-\frac{12(1^2-4)}{1^4}=+$
$f'(3)=-\frac{12(3^2-4)}{3^4}=-$
The positive values is where $f(x)$ is increasing.
The interval where $f(x)$ is increasing is (0,2).
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(6) Find the $x$-value(s) of all local maxima. (Find exact values, and not decimal representations)
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(6)In order to find local maxima, the derivative of the slope is used. The derivative of the slope is
$f''(x)=\frac{24(x^2-8)}{x^5}$
Plug in the critical points found in an earlier question. If the value is less than zero, that is a local maximum.
$f''(0)=\frac{24(0^2-8)}{0^5}=-192$
$f''(2)=\frac{24(2^2-8)}{2^5}=-3$
$f''(-2)=\frac{24(-2^2-8)}{-2^5}=9$
The x-values of all local maxima are 0 and 2.

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(7) Find the $x$-value(s) of all local minima. (Find exact values, and not decimal representations)
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(7)In order to find local minima, the derivative of the slope is used. The derivative of the slope is
$f''(x)=\frac{24(x^2-8)}{x^5}$
Plug in the critical points found in an earlier question. If the value is greater than zero, that is a local minimum.
$f''(0)=\frac{24(0^2-8)}{0^5}=-192$
$f''(2)=\frac{24(2^2-8)}{2^5}=-3$
$f''(-2)=\frac{24(-2^2-8)}{-2^5}=9$
The x-value of local minima is -2.

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(8) Find the interval(s) on which the graph is concave downward.
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(8)In order to find which interval(s) are concave downward, the second derivative is used. The second derivative is $f''(x)=\frac{24(x^2-8)}{x^5}$
Set the second derivative to equal zero
$f''(x)=\frac{24(x^2-8)}{x^5}=0$
The critical points for the second derivative are $x=0,2\sqrt2,-2\sqrt2$
The critical points are placed on a number line and integars above the critical points are plugged into the second derivative of the function to find the sign (positive or negative) of the derivative of the slope.
The numbers I choose to plug into the second derivative are -3,-1,1,and 3.
$f''(x)=\frac{24(x^2-8)}{x^5}$
$f''(-3)=\frac{24(-3^2-8)}{-3^5}=+$
$f''(-1)=\frac{24(-1^2-8)}{-1^5}=+$
$f''(1)=\frac{24(1^2-8)}{1^5}=-$
$f''(3)=\frac{24(3^2-8)}{3^5}=+$
When the values are negative that makes $f(x)$ concave down.
The interval on which the graph is concave down is $(0,2\sqrt2)$

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(9) State the $x$-value(s) of all inflection points. (Find exact values, and not decimal representations)
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(9)In order to find the inflection points, the sign chart that was made in the earlier question is referred to. The numbers I choose to plug into the second derivative are -3,-1,1,and 3.
$f''(x)=\frac{24(x^2-8)}{x^5}$
$f''(-3)=\frac{24(-3^2-8)}{-3^5}=+$
$f''(-1)=\frac{24(-1^2-8)}{-1^5}=+$
$f''(1)=\frac{24(1^2-8)}{1^5}=-$
$f''(3)=\frac{24(3^2-8)}{3^5}=+$
When there is a sign change in the sign chart, those are the points of inflection. In this case, the x-values of all inflection points are $0,2\sqrt2,-2\sqrt2$.

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(10) Include a sketch of the graph of $y=f(x)$. Plot the different segments of the graph using the color code below.
* **blue:** $f'>0$ and $f''>0$
* **red:** $f'<0$ and $f''>0$
* **black:** $f'>0$ and $f''<0$
* **gold:** $f'<0$ and $f''<0$

(In Desmos you could restrict the plot $y=f(x)$ on the interval $[2,3]$ by typing $y=f(x)\{2\le x\le 3\}$.) Be sure to set the bounds on the graph so that the features of the graph that you listed above are easy to see.
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(10)
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