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)$f(x)=\frac{12x^2-16}{x^3}$ the domain is $(-infinity,0)U(0,infinity)$
$f'(x)=\frac{12(x^2-4)}{x^4}$ domain is $(-infinity,0)U(0,infinity)$
$f''(x)=\frac{24(x^2-8)}{x^5}$ domain is $(-infinity,0)U(0,infinity)$

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(2) Find all $x$- and $y$-intercepts.
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(2)$f(x)=\frac{12(x^2-16)}{x^3}$ intercept X $(-\frac{2sqrt(3)}{3},0)(\frac{2sqrt(3)}{3},0)$and y has no intercepts
$f'(x)=\frac{12(x^2-4)}{x^4}$ intercept X $(2,0)(-2,0)$ and y has no intercpts
$f''(x)=\frac{24(x^2-8)}{x^5}$ intercept x $(2sqrt2,0),(-2sqrt2,0)$ and y has no intercepts

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(3) Find all equations of horizontal asymptotes.
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(3)$f(x)=\frac{12(x^2-16)}{x^3}$ y=0
$f'(x)=\frac{12(x^2-4)}{x^4}$ y=0
$f''(x)=\frac{24(x^2-8)}{x^5}$ y=0

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(4) Find all equations of vertical asymptotes.
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(4) $f(x)=\frac{12(x^2-4)}{x^3}$ x=0
$f'(x)=\frac{12(x^2-4)}{x^4}$ x=0
$f''(x)\frac{24(x^2-8)}{x^5}$ x=0

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(5) Find the interval(s) where $f$ is increasing.
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(5)$f(x)=\frac{12(x^2-16)}{x^3}$ increases on $(-2,0),(0,2)$
$f'(x)=\frac{12(x^2-4)}{x^4}$ increases on $(-infinity,-2sqrt2),(0,2sqrt2)$
$f''(x)=\frac{24(x^2-8)}{x^5}$ increases on $(--\frac{2sqrt30}{3},0),(0,\frac{2sqrt30}{3})$

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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)$f(x)=\frac{12(x^2-16)}{x^3}$ is $f(x)=\frac{4(3x^2-4)}{x^3}$
$f'(x)=\frac{12(x^2-4)}{x^4}$ is $f'(x)=\frac{12(x+2)(x-2)}{x^4}$
$f''(x)=\frac{24(x^2-8)}{x^5}$ is $f''(x)=\frac{24x^2-192}{x^5}$

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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)$f(x)=\frac{12(x^2-16)}{x^3}$ is $(-2,-4)$
$f'(x)=\frac{12(x^2-4)}{x^4}$ is $(2sqrt2,\frac{3}{4})$

$f''(x)=\frac{24(x^2-8)}{x^5}$ is $(-\frac{2sqrt30}{3},-\frac {9sqrt30}{250})$

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(8) Find the interval(s) on which the graph is concave downward.
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(8)$f(x)=\frac{12(x^2-16)}{x^3}$ are $(infinity,-2sqrt2)$ and $(0,2sqrt2)$
$f'(x)=\frac{12(x^2-4)}{x^4}$ are $(-\frac{2sqrt30}{3},0)$and $(0,\frac{2sqrt30}{3})$
$f''(x)=\frac{24(x^2-8)}{x^5}$ are $(-infinity,-2sqrt5)$ and $(0,2sqrt5)$

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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)$f(x)=\frac{12(x^2-16)}{x^3}$ is $(2sqrt2,\frac{5sqrt2}{2})$
$f'(x)=\frac{12(x^2-4)}{x^4}$ is $(-\frac{2sqrt30}{3},\frac{63}{100})$ and $\frac{2sqrt30}{3},\frac{63}{100}$
$f''(x)=\frac{24(x^2-8)}{x^5}$ is $(2sqrt5,\frac{9sqrt5}{125})$

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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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