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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 
$$
f(x)=\frac{12x^2-16}{x^3},\qquad f'(x)=-\frac{12(x^2-4)}{x^4},\qquad f''(![](https://i.imgur.com/zte82bw.png)
x)=\frac{24(x^2-8)}{x^5}.
$$ 
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)
All except x=0
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(2) Find all $x$- and $y$-intercepts. 
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(2) Set f(x)=0 and Solve for x-intercept
$0=(12x^2-16)$/$(x^3)$
$x^3+16/12$=$x^2$
$\left(\sqrt{x^{3}+4}\div3\right)=x^{2}$
=+-$2\sqrt{3}\div3=x$ or $+-1.54$
replaced x with 0 to solve for y-intercept and y-intercept=0
So x- and y intercepts are $(+-1.54,0)$ .




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(3) Find all equations of horizontal asymptotes. 
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The degree of the denominator is greater than the degree of numerator  so the equation for the horizontal asymptotes is $y=0$
(3)

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(4) Find all equations of vertical asymptotes. 
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(4)Since the function is defined everywhere except where x=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)
1st I found critical Value of $f'(x)=-12(x^2-4)/(x^4)$
    So I set f'(x)=0 
    $(x-2)=+2$
    $(x+2)=-2$
    f'(x) is defined everywhere except where it equals to zero so x=0 is another critical point. 
    So my critical points are $x=2,-2,0$
  Next I used the 1st Derivative Test. In which I plotted my critical point on the real number line and selected test value in between my intervals. I concluded that  $f$ is increasing in intervals
  $(-2,0) and (0,2)$
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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)$f$ has a Local maxima $f(2)=(12-(2)^2-16) /(2)^3=4$

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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)$f$ has a Local minima $f(-2)=(12-(-2)^2-16) /(-2)^3=-4$

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(8) Find the interval(s) on which the graph is concave downward. 
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(8)1st I found the critical values for $f''(x)=24(x^2-8) /x^5$
Set f''(x)=0
$(x^2-8)=+- $2\sqrt{2}$ or +-2.83$
NOte that f''(x) is defined everywhere except where it equals to zero so x=0 is another critical point
So my critical values for f''(x) are $x=-2.83,2.83,0$
Then I used the 2nd Derivative  In which I plotted my critical point on the real number line and selected test value in between my intervals. I concluded that the graph is concave downward in the intervals $(-2.83,0)$ and $(-inf,-2.83)$ and the graph is concave up $(0,2.83)and (2.83, inf)$
> 
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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($2\sqrt{2}$)=$\frac{12(2\sqrt{2})^2-16}{(2\sqrt{2})^3}$=$(96-16)/(8)(\sqrt{2})^3)$= $\frac{10}{(\sqrt{2})^3}$
f($-2\sqrt{2}$)=$\frac{12(-2\sqrt{2})^2-16}{(-2\sqrt{2})^3}$=$(96-16)/(8)(\sqrt{2})^3)$= $-\frac{10}{(\sqrt{2})^3}$

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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/TWqOQyo.png)







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To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.



