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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 $f(x)$ should include all real numbers, exclusive of $0$. The function $f(x)$ is undefined when $x=0$ since that would result in having $0$ in the denominator, which causes $f(x)$ to be undefined at this value of $x$. 
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(2) Find all $x$- and $y$-intercepts. 
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(2)
$x$-intercepts:
Set the numerator of $f(x)=0$ and then solve for $x$:

$(12x^2-16)=0$

$x^2=\frac{16}{12}$

$x=\sqrt(\frac{16}{12})$

$x=\frac{4}{\sqrt(12)}$, $\frac{-4}{\sqrt(12)}$

$y$-intercept: 

$f(0)=\frac{12(0)^2-16}{0^3}$

There would be no $y$-intercept for $f(x)$ because $f(x)$ is undefined when $x=0$ (the input value when a function would cross the $y$-axis)


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(3) Find all equations of horizontal asymptotes. 
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(3)
Since the highest degree of the polynomial in the numerator is 2 and this number is smaller than the highest degree of the polynomial in the denominator (which is 3), the formula for the horizontal asymptote for $f(x)$ is $y=0$ (it is the $x$-axis).
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(4) Find all equations of vertical asymptotes. 
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(4)
The vertical asymptote of $f(x)$ would be located at any $x$-value at which $f(x)$ is undefined. Looking at the solution provided to question 1, there should be a vertical asymptote at $x=0$ since plugging in $0$ into the denominator causes $f(x)$ to be undefined.
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(5) Find the interval(s) where $f$ is increasing.
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(5)
Finding the critical values by setting $f'(x)=0$: 

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

$f'(x)$ is undefined when $x=0$ (zero in the denominator)

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

$(x^2-4)=0$

$x=-2, +2$

The critical values of $f'(x)$ are $x=-2,0,+2$

Conduct the first derivative test using selected values between the determined critical values:

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

$f'(-100)=\frac{-12((-100)^2-4)}{(-100)^4}=-1.2x10^{-3}$
$f(x)$ is decreasing when $x<-2$

$f'(-1)=\frac{-12((-1)^2-4)}{(-1)^4}=+36$
$f(x)$ is increasing when $-2<x<0$

$f'(100)=\frac{-12((100)^2-4)}{(100)^4}=-1.2x10^{-3}$
$f(x)$ is decreasing when $x>2$

The function $f(x)$ is increasing only when $-2<x<0$.

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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)
From conducting the First Derivative Test in the previous question:

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

$f'(-100)=\frac{-12((-100)^2-4)}{(-100)^4}=-1.2x10^{-3}$
$f(x)$ is decreasing when $x<-2$

$f'(-1)=\frac{-12((-1)^2-4)}{(-1)^4}=+36$
$f(x)$ is increasing when $-2<x<0$

$f'(100)=\frac{-12((100)^2-4)}{(100)^4}=-1.2x10^{-3}$
$f(x)$ is decreasing when $x>2$

The function $f(x)$ has a local maxima at $x=2$ since the function was shown to be increasing prior to this critical $x$-value and then decreasing following this $x$-value.
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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)
From conducting the First Derivative Test in the previous question:

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

$f'(-100)=\frac{-12((-100)^2-4)}{(-100)^4}=-1.2x10^{-3}$
$f(x)$ is decreasing when $x<-2$

$f'(-1)=\frac{-12((-1)^2-4)}{(-1)^4}=+36$
$f(x)$ is increasing when $-2<x<0$

$f'(100)=\frac{-12((100)^2-4)}{(100)^4}=-1.2x10^{-3}$
$f(x)$ is decreasing when $x>2$

The function $f(x)$ has a local minima at $x=-2$ since the function was shown to be decreasing prior to this critical $x$-value and then increasing following this $x$-value.

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(8) Find the interval(s) on which the graph is concave downward. 
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(8)
Finding the critical values of $f"(x)$:

Given: $f"(x)=\frac{24(x^2-8)}{x^5}$

$f"(x)$ is undefined when $x=0$ since plugging zero into the denominator results in an expression involving dividing a polynomial by zero. 

Setting the numerator equal to zero then solving for $x$:

$24(x^2-8)=0$

$x^2-8=0$

$x^2-8=0$

$x^2=8$

$x=\sqrt8$ is a critical value for $f"(x)$

Using selected values from the intervals determined by the two calculated critical values of $f"(x)$ for determining the concavity of $f(x)$:

Given: $f"(x)=\frac{24(x^2-8)}{x^5}$

$f"(-10)=\frac{24((-10)^2-8)}{(-10)^5}=+0.022$
$f(x)$ is concave up for $x$-values $<0$.

$f"(1)=\frac{24((1)^2-8)}{(1)^5}=-192$
$f(x)$ is concave down for $x$-values $0<x<\sqrt8$.

$f"(100)=\frac{24((100)^2-8)}{(100)^5}=2.4x10^{-5}$
$f(x)$ is concave up for $x$-values $> \sqrt8$.

The function $f(x)$ is concave downward for $x$-values $0<x<\sqrt8$.

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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 work from the test conducted for the previous question: 

Given: $f"(x)=\frac{24(x^2-8)}{x^5}$

$f"(-10)=\frac{24((-10)^2-8)}{(-10)^5}=+0.022$
$f(x)$ is concave up for $x$-values $<0$.

$f"(1)=\frac{24((1)^2-8)}{(1)^5}=-192$
$f(x)$ is concave down for $x$-values $0<x<\sqrt8$.

$f"(100)=\frac{24((100)^2-8)}{(100)^5}=2.4x10^{-5}$
$f(x)$ is concave up for $x$-values $> \sqrt8$.

The function $f(x)$ is concave downward for $x$-values $0<x<\sqrt8$.

The inflection points of $f(x)$ are values in which the function changes in concavity. The inflection points are $x=0, \sqrt8$.

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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)
Blue: $f(x)$ is concave up and has a positive slope
Red: $f(x)$ is concave up and has a negative slope
Black: $f(x)$ is concave down and has a positive slope
Gold: $f(x)$ is concave down and has a negative slope

![](https://i.imgur.com/EwMdQjZ.png)








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