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)
$(-\infty,0]$ and $[0,\infty)$
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(2) Find all $x$- and $y$-intercepts.
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(2)
$\frac{12x^2-16}{x^3}=0$
$12x^2=16$
$x^2=16/12$
$x^2=4/3$
$x= + or - \sqrt(\frac{4}{3})$
$y=\frac{12(0)^2-16}{(0)^3}$
$y=-16/0$
y is undefined

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(3) Find all equations of horizontal asymptotes.
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(3)
y=0 since f(x) never touches 0 at y
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(4) Find all equations of vertical asymptotes.
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(4)
x=0 since f(x) never touches 0 at x
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(5) Find the interval(s) where $f$ is increasing.
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(5)
$\frac{-12(x^2-4)}{x^4}=0$
$-12(x^2-4)=0$
$x^2-4=0$
$x^2=4$
$x=+ or - 2$
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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)
$f'(x)=-\frac{12(x^2-4)}{x^4}$
f'(1)>0
f'(3)<0
x=2 local maxima
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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)
$f'(x)=-\frac{12(x^2-4)}{x^4}$
f'(-3)<0
f'(-1)>0
x=-2 local minima
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(8) Find the interval(s) on which the graph is concave downward.
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(8)
$f''(x)=\frac{24(x^2-8)}{x^5}$
$24(x^2-8)=0$
$x^2=8$
$x=+ or - \sqrt{8}$
f''($\sqrt{8}$)= (+) and concave up
f''($-\sqrt{8}$)= (-) and concave down

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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)
$f''(x)=\frac{24(x^2-8)}{x^5}$
$\frac{24(x^2-8)}{x^5}=0$
$24(x^2-8)=0$
$x^2=8$
$x=+ or - \sqrt{8}$
f''(-3)<0
f''($-\sqrt{8}$) inflection point
f''(-1)>0
f''(1)<0
f''($\sqrt{8}$) inflection point
f''(3)>0

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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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