> []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) The domain of the functions are:
1.$f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ the domain is $\left(-\infty,0\right)U\left(0,\infty\right)$
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ the domain is also $\left(-\infty,0\right)U\left(0,\infty\right)$
3.$f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ the domain again is $\left(-\infty,0\right)U\left(0,\infty\right)$
we do this by setting the domain to $x^{3}$$\ne$ 0

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(2) Find all $x$- and $y$-intercepts.
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(2)The x and y intercepts for these are
1.$f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ they are $\left(-\frac{2\sqrt{3}}{3},0\right)\left(\frac{2\sqrt{3}}{3},0\right)$ for the x value and for y there are none.
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ they are (2,0),(-2,0) for the x value and for y there are also none.
3.$f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ they are $\left(2\sqrt{2},0\right),\left(-2\sqrt{2},0\right)$ forthe x values and for y there are again none.
we do this by setting the numerator to 0 and for the y-values the limit doesnt exsist so we show it as $\lim_{x\to\infty}$

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(3) Find all equations of horizontal asymptotes.
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(3) The horizontal asymptotes are:
1.$f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ it is y=0
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ it is y=0
3.$f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ it is y=0
we observe that degree of the numeraator = 2 and for the denominator it = 3 so 2<3
this being the degree of the numerator<degree of the denominator that being shown as y=0 and that explains why y=0 for those horizontal asymptotes

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(4) Find all equations of vertical asymptotes.
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(4) The vertical asymptotes are:
1.$f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ it is x=0
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ it is x=0
3.$f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ it is x=0
for this we set the denominator to 0 and that is shown as x^3=0 showing why it is x=0 for the vertical asymptotes

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

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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 for the x values are:
1.$f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ it is $f\left(x\right)=\frac{4\left(3x^{2}-4\right)}{x^{3}}$
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ it is $f'\left(x\right)=\frac{12\left(x+2\right)\left(x-2\right)}{x^{4}}$
3.$f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ it 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 local minima are:
1.$f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ it is (-2,-4)
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ it is $\left(2\sqrt{2},\frac{3}{4}\right)$
3.$f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ it 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 it is concave downwards are:
1.$f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ they are $\left(-\infty,-2\sqrt{2}\right)$ and $\left(0,2\sqrt{2}\right)$
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ they are $\left(-\frac{2\sqrt{30}}{3},0\right)$ and $\left(0,\frac{2\sqrt{30}}{3}\right)$
3.$f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ they are $\left(-\infty,-2\sqrt{5}\right)$ 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.$f\left(x\right)=\frac{12\left(x^{2}-16\right)}{x^{3}}$ it is $\left(2\sqrt{2},\frac{5\sqrt{2}}{2}\right)$
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ they are $\left(-\frac{2\sqrt{30}}{3},\frac{63}{100}\right)$ and $\left(\frac{2\sqrt{30}}{3},\ \frac{63}{100}\right)$
3.$f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ it 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)

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