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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)
The domain of the function $f$ is all real numbers except for 0.

$(-infinity,0)U(0,infinity)$

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(2) Find all $x$- and $y$-intercepts. 
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To get the $x$- intercepts, set $y=0$ and solve:
$0=\frac{12x^2-16}{x^3}$
$0=12x^2-16$
$16=12x^2$
$\frac{16}{12}=x^2$
$\frac{+}{-}\sqrt\frac{16}{12}=x$
$x=\frac{2\sqrt3}{3}$
$x=-\frac{2\sqrt3}{3}$

The $x$- intercepts for $f(x)$ are $\frac{2\sqrt3}{3}$ and $-\frac{2\sqrt3}{3}$  .
When you set $x=0$ to find the $y$- intercepts, the denominator becomes $0$ making the equation undefined, therefore there are no $y$- intercepts.

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(3) Find all equations of horizontal asymptotes. 
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(3)To find the horizontal asymptotes, look at the degree of the leading coefficients for $f(x)$:
$f(x)=\frac{12x^2-16}{x^3}$
The numerator is degree 2 and the denominator is degree 3, therefore the horizontal asymptote is $y=0$. 

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(4) Find all equations of vertical asymptotes. 
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(4) To find the vertical asymptote(s), we can take the limit of the function as it moves closer to $-infinity$ and $infinity$ using L'Hopital's Rule:
$f(x)=\lim_{h \to infinity}\frac{12x^2-16}{x^3}$
$f(x)\frac{LH}{=}\lim_{h \to infinity}\frac{24x}{3x^2}$
$f(x)\frac{LH}{=}\lim_{h \to infinity}\frac{24}{6x}$
$f(x)\frac{LH}{=}\lim_{h \to infinity}\frac{0}{6}= 0$

The equation for the vertical asymptote is $x=0$.



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(5) Find the interval(s) where $f$ is increasing.
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(5)Using the derivative of $f(x)$ stated above:
$0=-\frac{12(x^2-4)}{x^4}$
$0=-12(x^2-4)$
$0=x^2-4$
$4=x^2$
$\frac{+}{-} \sqrt4 =x$
$x=- 2$
$x=2$

During the interval $x<-2$, $f(x)$ is decreasing.
During the interval $-2<x<2$, $f(x)$ is increasing.
During the interval $x>2$, $f(x)$ is decreasing.

The interval where $f(x)$ is increasing is $(-2,2)$.




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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)Using the first derivative test from #5, the $x$ value of the local maxima is $x=2$.

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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) Using the first derivative test from #5, the $x$ value of the local minima is $x=-2$.

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(8) Find the interval(s) on which the graph is concave downward. 
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(8)$f''(x)=\frac{24(x^2-8)}{x^5}$
$0=\frac{24(x^2-8)}{x^5}$
$0=24(x^2-8)$
$0=x^2-8$
$8=x^2$
$x=\frac{+}{-}\sqrt8$
$x=\frac{+}{-}2\sqrt2$
On the interval $x<-2\sqrt2$,  $f''(x)=-$ therefore $f(x)=$ concave down.
On the interval $-2\sqrt2<x<0$,  $f''(x)=+$ therefore $f(x)=$ concave up.
On the interval $0<x<2\sqrt2$,  $f''(x)=-$ therefore $f(x)=$ concave down.
On the interval $2\sqrt2<x$,  $f''(x)=+$ therefore $f(x)=$ concave up.

So, the intervals where $f(x)$ is concave down are $(-infinity, -2\sqrt2)U(0,2\sqrt2)$.


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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)
$f(-2\sqrt2)=\frac{12(-2\sqrt2)^2-16}{(-2\sqrt2)^3}$
$f(-2\sqrt2)=\frac{12(4\cdot2)-16}{(-8\cdot2\cdot\sqrt2)}$
$f(-2\sqrt2)=\frac{12(8)-16}{-16\sqrt2}$
$f(-2\sqrt2)=\frac{80}{-16\sqrt2}$
$f(-2\sqrt2)=-\frac{5}{\sqrt2}$

$f(2\sqrt2)=\frac{12(2\sqrt2)^2-16}{(2\sqrt2)^3}$
$f(2\sqrt2)=\frac{12(4\cdot2)-16}{(8\cdot2\cdot\sqrt2)}$
$f(2\sqrt2)=\frac{12(8)-16}{16\sqrt2}$
$f(2\sqrt2)=\frac{80}{16\sqrt2}$
$f(2\sqrt2)=\frac{5}{\sqrt2}$




The inflection points for $f(x)$ are $(-2\sqrt2,-\frac{5}{\sqrt2})$ and $(2\sqrt2,\frac{5}{\sqrt2})$.

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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/cLZAtkB.png)







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