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) The domain of f(X) is (-∞, 0)(0, ∞)

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(2) Find all $x$- and $y$-intercepts.
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(2)
x-intercepts
$f(x)=\frac{12x^2-16}{x^3}$
Set f equal to zero and solve.
$\frac{12x^2-16}{x^3}=0$
$12x^2-16$
$x=-

y-intercept
When we set $x^4=0$ it turns out that x=0 is a vertical asymptote.

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(3) Find all equations of horizontal asymptotes.
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(3)
$f(x)=\frac{12x^2-16}{x^3}$
Using the rules of horizontal asymptotes, we can conclude that the horizontal asymptote is y=0. This is because the degree of the denominator is bigger than the numerator.

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(4) Find all equations of vertical asymptotes.
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(4)
When we set the denominator equal to zero we get x=0, which is the vertical asymptote. The denomitor can't literally equal 0 because it'll make the function undefined.

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(5) Find the interval(s) where $f$ is increasing.
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(5)
f is increasing on the interval (-∞, -2) and (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, we can conclude:
Using our number line we can see that the local maximum is when the sign of f' switches from positive to negative. Therefore, the maximum is located at 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 our number line we can see that the local minimum is when the sign of f' switches from negative to positive. Therefore, the maximum is located at x=2.

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(8) Find the interval(s) on which the graph is concave downward.
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(8)
Set f'' equal to zero.
$f''(x)=\frac{24(x^2-8)}{x^5}=0$
$x^2-8=0$ and $x^5=0$
Our critical values are $x=-2\sqrt{2}, 2\sqrt{2}$

Then we follow the rules of the second derivative test and we can conclude:


Therefore, f is concave down on the intervals $(-∞, -2\sqrt{2})$ and $(0, 2\sqrt{2})$.

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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)
Using the line segment from the last problem we can conclude that our inflection points are located at $x=-2\sqrt{2}, 0, 2\sqrt{2}$. This is because inflection points are located wherever there is a change in concavity.

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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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