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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)

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(2) Find all $x$- and $y$-intercepts. 
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(2)$$f(x)=\frac{12x^2-16}{x^3}$$
$$y=\frac{12(0)^2-16}{(0)^3}$$
$$y=undefined$$

$$
\frac{12x^2-16}{x^3}=0$$
$$
\frac{12x^2-16}{x^3 *x^3}=0*x^3$$
$$
12x^2-16=0$$
$$12x^2-16+16=0+16$$
$$\frac{12x^2}{12}=\frac{16}{12}$$
$$\sqrt{x^2}=\sqrt{ \frac{4}{3}}$$
$$x=1.1547, x=-1.1547$$





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(3) Find all equations of horizontal asymptotes. 
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(3)$$f(x)=\frac{12x^2-16}{x^3}$$
lim $x→∞=f(x)$
Apply L'Hopital's rule
lim $x→∞ \frac{12x^2-16}{x^3}$ =lim $x→∞ \frac{\frac{d}{dx}({12x^2-16)}}{\frac{d}{dx}{(x^3)}}=$lim $x→∞ \frac{24x}{3x^2}$ 
Apply L'Hopitals rule again
lim $x→∞ \frac{\frac{d}{dx}(24x)}{\frac{d}{dx}{(3x^2)}}$=lim $x→∞ \frac{24}{6x}$ 
Once more (to make sure there isn no divison by zero)
lim $x→∞ \frac{24}{6x}$=lim $x→∞ \frac{\frac{d}{dx}(24)}{\frac{d}{dx}{(6x)}}$ =lim $x→∞ \frac{0}{6}$ =0
The horizontal asymptote is zero 



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(4) Find all equations of vertical asymptotes. 
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(4)$$f(x)=\frac{12x^2-16}{x^3}$$
$${x^3}=0$$
$$x=0$$
the verticical asymptote is zero 


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(5) Find the interval(s) where $f$ is increasing.
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(5) Noting the domain, $x^3=0 → x=0$ I applied the 1st derivative test where:
$$f'(-1)=36 ,  f'(1)=36$$
both postive values hence the intervals  where f(x) is increasing are:
$$(-∞,0)U(0,∞)$$

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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)

$f'(x)=-\frac{12(x^2-4)}{x^4}$

$\frac{12(x^2-4)}{x^4*x^4}=0*x^4$

$12(x^2-4)=0$

$\frac{-12(x^2-4)}{12}=\frac{0}{12}$

$x^2-4+4=0+4$

$x^2=4$

$\sqrt{ x^2}=\sqrt{ 4}$

$x=2, x=-2$

f''(2)<0, f''(-2)>0
local maxima where: $f(2)=\frac{12(2)^2-16}{(2)^3}=4$
because f''(2)<0
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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)*Based on work from #6 there is a local minimum where: 
$$f(-2)=\frac{12(-2)^2-16}{(-2)^3}=-4$$

because f''(-2)>0

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(8) Find the interval(s) on which the graph is concave downward. 
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(8) Applying the 2nd derivative test noting the domain is x=0 where f''(-1)=168 and f''(1)=-168 . the intervals where f(x) is concave down is:
$$(0,∞)$$
because f''(1)<0
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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) $$f''(x)=\frac{24(x^2-8)}{x^5}$$
$$\frac{24(x^2-8)}{x^5}=0$$
$$\frac{24(x^2-8)}{x^5*x^5}=0*x^5$$
$$24(x^2-8)=0$$
$$\frac{24(x^2-8)}{24}=\frac{0}{24}$$
$$x^2-8+8=0+8$$
$$\sqrt{x^2}=\sqrt8$$
$$x=\sqrt{4*2}$$
$$x=\sqrt{4}\sqrt2$$
the x values  of the inflections points are
$$x=2\sqrt{2},- 2\sqrt{2}$$





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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/4NgiBcp.png)







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