Math 181 Miniproject 7: The Shape of a Graph.md
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tags: MATH 181
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Math 181 Miniproject 7: The Shape of a Graph
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**Overview:** In this miniproject you will be using the techniques of calculus to find the behavior of a graph.

**Prerequisites:** The project draws heavily from the ideas of Chapter 1 and $2.8$ together with ideas and techniques of the first and second derivative tests from $3.1$.

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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)$f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$
To find the domain of the given function, set $x^3≠0$
$x≠0$
$(-∞,0)U(0,∞)$
The domain of the given function is $(-∞,0)U(0,∞)$

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(2) Find all $x$- and $y$-intercepts.
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(2)$f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$
to find the x-intercepts, set the numerator=0
$12x^{2}-16=0$
$12x^{2}=16$
$x=\pm\frac{4}{\sqrt{3}}$
The x-intercepts $(-\frac{4}{\sqrt{3}},0)$ and $(\frac{4}{\sqrt{3}},0)$
To find the y-intercepts, set x =0 in the given function
$f\left(0\right)=-\frac{16}{0}$
So, y-intercepts DNE

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(3) Find all equations of horizontal asymptotes.
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(3)$f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$
from the function, the degree of the numerator is 2 and the degree of the denominator is 3 so 2<3. That degree is the degree of the numerator less than the degree of the denominator, which means the equation of the horizontal asymptote is y=0

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(4) Find all equations of vertical asymptotes.
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(4)$f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$
To find the verticle asymptotes of the given function, set the denominator equal to 0 that is $x^3=0$
So the equation of the asymptote is x=0

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(5) Find the interval(s) where $f$ is increasing.
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(5) The intervals where x is increasing are
1. $f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ it's increasing on(-2,0),(0,2)
2. $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ it's increasing on $(-∞,2\sqrt{2}),(0,2\sqrt{2})$
3. $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ it's increasing on $(\frac{-2\sqrt{30}}{3},0),\left(0,\frac{2\sqrt{30}}{3}\right)$

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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) The exact values of x are
1. $f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ is $f\left(x\right)=\frac{4\left(3x^{2}-4\right)}{x^{3}}$
2. $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ is $f'\left(x\right)=\frac{12\left(x+2\right)\left(x-2\right)}{x^{4}}$
3. $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ is $f''\left(x\right)=\frac{24x^{2}-192}{x^{5}}$

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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) The x values of the local minima are
1. $f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ is (-2,-4)
2. $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ is $\left(2\sqrt{2},\frac{3}{4}\right)$
3. $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ is $\left(\frac{-2\sqrt{30}}{3},\frac{-9\sqrt{30}}{250}\right)$

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(8) Find the interval(s) on which the graph is concave downward.
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(8) The intervals where the concave is downward are
1. $f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ are $\left(-∞,-2\sqrt{2}\right)\ and\ \left(0,2\sqrt{2}\right)$
2. $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ are $\left(-\frac{2\sqrt{30}}{3},0\right)\ and\ \left(0,\frac{2\sqrt{30}}{3}\right)$
3. $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ are $\left(-∞,-2\sqrt{5}\right)\ and\ \left(0,2\sqrt{5}\right)$

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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) All the x values inflection points are
1. $f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ is $\left(2\sqrt{2},\frac{5\sqrt{2}}{2}\right)$
2. $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ are $\left(-\frac{2\sqrt{30}}{3},\frac{63}{100}\right)\ and\ \left(\frac{2\sqrt{30}}{3},\frac{63}{100}\right)$
3. $f''\left(x\right)=\frac{24\left(x-8\right)}{x^{5}}\ is\ \left(2\sqrt{5},\frac{9\sqrt{5}}{125}\right)$

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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)<iframe src="https://www.desmos.com/calculator/murjnr3zfm?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

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