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title: 'Math 181 Miniproject 7: The Shape of a Graph'

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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
===

**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) 1. $f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ DOMAIN: $\left(-\infty,0\right)U\left(0,\infty\right)$
2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ DOMAIN: $\left(-\infty,0\right)U\left(0,\infty\right)$
3.3. $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ DOMAIN: $\left(-\infty,0\right)U\left(0,\infty\right)$
So basically the same domain for all of the functions. 

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(2) Find all $x$- and $y$-intercepts. 
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(2) To find the x-values all you do is plug in 0 for y and then solve for x. To find the y-intercept you would plug in 0 for x and solve for y. Then repeating these steps to find the domain of the other functions.  
1. x value: $\left(-\frac{2\sqrt{\left(3\right)}}{3},0\right),\left(\frac{2\sqrt{\left(3\right)}}{3},0\right)$ and none for the y value
2.x value: (2,0), (-2,0) and none for the y value
3.x value: $\left(2\sqrt{\left(2\right)},0\right),\ \left(-2\sqrt{2},0\right)$ and none for the y value


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(3) Find all equations of horizontal asymptotes. 
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(3) 1. for the fucntion $f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ the horizontal asymptote is y = 0.
2. For the fucntion $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ the horizontal asymptote is y = 0.
3. 3. For the function $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ the horizontal asymptote is y = 0.

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(4) Find all equations of vertical asymptotes. 
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(4) 1. For the function $f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ the vertical asymptote is x = 0.
2. For the fucntion $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ the vertical asymptote is x = 0.
3. For the function $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ the vertical asymptote is x = 0.

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(5) Find the interval(s) where $f$ is increasing.
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(5) 1. $f(x)$ is increasing (-2,0),(0,2)
2.$f'(x)$ is increasing $\left(-\infty,2\sqrt{\left(2\right)}\right),\left(0,2\sqrt{\left(2\right)}\right)$
3.$f''(x)$ is increasing $\left(\frac{-2\sqrt{\left(30\right)}}{3},0\right),\left(0,\frac{2\sqrt{\left(30\right)}}{3}\right)$

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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) To find the local maxima you would first find where the first derivative is equal to 0.
1. The x-value of the local maxima for the function $f(x)=f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ is 2. 
2. There is no local maxima for the function $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$. 
3. The x-value for the local maxima for the function $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ is $\frac{2\sqrt{\left(30\right)}}{3}$.

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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) To find the local minima you would first find where the first derivative is equal to 0. 
1. The x-value of the local minima for $f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ is -2.
2. The local minima for the function $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ is $2\sqrt{\left(2\right)}$.  
3. The local minima for the function $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ is $\frac{-2\sqrt{\left(30\right)}}{3}$.

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(8) Find the interval(s) on which the graph is concave downward. 
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(8) 1. The graph is concave down for the graph $f(x)$ at $\left(-\infty,-2\sqrt{\left(2\right)}\right)and\ \left(0,2\sqrt{\left(2\right)}\right)$
2. The graph is concave down for the graph $f''(x)$ at $\left(\frac{-2\sqrt{\left(30\right)}}{3},0\right)and\left(0,\frac{2\sqrt{\left(30\right)}}{3}\right)$
3. The graph is concave down for the graph $f''(x)$ at $\left(-\infty,-2\sqrt{\left(5\right)}\right)and\left(0,2\sqrt{\left(5\right)}\right)$

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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) To find the inflection points, you would first find where the second derivative is equal to 0. 
1. Inflection point x-value for $f(x)$ is $2\sqrt{\left(2\right)}$
2. There are no inflection points for $f'(x)$ 
3. Inflection point x-value for $f''(x)$ is $2\sqrt{\left(5\right)}$ 
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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)
![](https://i.imgur.com/TODUtlO.png)






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