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 is all real numbers expect when x=0.

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(2) Find all $x$- and $y$-intercepts.
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(2)
X-intercept
$f(x) = \frac{\left(12x^{2\ }-\ 16\right)}{x^{3}}$

$12x^{2}\ -\ 16\ =\ 0$
$12x^{2}=16$
$x^{2\ }=\frac{16}{12}$
$x=\sqrt{\frac{16}{12}}=+/-1.15470053838$
x-intercept = (-1.15,0), (1.15,0)

Y-intercept
$f(x) = \frac{\left(12x^{2\ }-\ 16\right)}{x^{3}}$
$\frac{\left(12\left(0\right)^{2}-16\right)}{0^{3}}$
y-intercept = undefined

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(3) Find all equations of horizontal asymptotes.
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(3)
The denominator of the function has a higher degree than the numerator. Therefore, the horizontal asymptote is at y = 0.

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(4) Find all equations of vertical asymptotes.
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(4)
$f(x) = \frac{\left(12x^{2\ }-\ 16\right)}{x^{3}}$

$12x^{2\ }-\ 16$
$4\left(3x^{2\ }-\ 4\right)$

The vertical asymptote is at x = 0.

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(5) Find the interval(s) where $f$ is increasing.
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(5)
$f'(x)=-\frac{12\left(x^{2\ }-4\right)}{x^{4}}$
$0=-\frac{12\left(x^{2\ }-4\right)}{x^{4}}$
$0=-12\left(x^{2}-4\right)$
$0=\left(x-2\right)\left(x+2\right)$
x=-2,2
f is increasing at the intervals of [-2,inf]

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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)
The critical values are at -2 and 2.
Test values at:
f'(-20)<0 the behavior of f'(-20) shows it is decreasing
f'(1)<0 the behavior of f'(1) shows it is decreasing
f'(20)>0 the behavior of f'(20) shows it is increasing

The local maxima is -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)
The local minimum is 2.

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(8) Find the interval(s) on which the graph is concave downward.
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(8)
$f"(x)=\frac{24\left(x^{2}-8\right)}{x^{5}}$

$0=\frac{24\left(x^{2}-8\right)}{x^{5}}$
$0=24\left(x^{2}-8\right)$
$0=x^{2}-8$
$8=x^{2}$
$+/-2.83=\sqrt{8}=\ x^{}$

Test values at:
f"(-10)<0 the behavior shows it would be decreasing
f"(0)<0 the behavior shows it would be decreasing
f"(10)>0 the behavior shows it would be increasing

The interval where it would be decreasing is $(-\infty,0)$.

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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)
The inflection points is at x = +/ - $\sqrt{8}$.

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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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