Math 181 Miniproject 7: The Shape of a Graph.md --- --- tags: MATH 181 --- 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 $3.1$. --- :::info 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. ::: (1) 1. $f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ DOMAIN: $\left(-\infty,0\right)U\left(0,\infty\right)$ 2.$f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ DOMAIN: $\left(-\infty,0\right)U\left(0,\infty\right)$ 3.3. $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ DOMAIN: $\left(-\infty,0\right)U\left(0,\infty\right)$ So basically the same domain for all of the functions. :::info (2) Find all $x$- and $y$-intercepts. ::: (2) To find the x-values all you do is plug in 0 for y and then solve for x. To find the y-intercept you would plug in 0 for x and solve for y. Then repeating these steps to find the domain of the other functions. 1. x value: $\left(-\frac{2\sqrt{\left(3\right)}}{3},0\right),\left(\frac{2\sqrt{\left(3\right)}}{3},0\right)$ and none for the y value 2.x value: (2,0), (-2,0) and none for the y value 3.x value: $\left(2\sqrt{\left(2\right)},0\right),\ \left(-2\sqrt{2},0\right)$ and none for the y value :::info (3) Find all equations of horizontal asymptotes. ::: (3) 1. for the fucntion $f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ the horizontal asymptote is y = 0. 2. For the fucntion $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ the horizontal asymptote is y = 0. 3. 3. For the function $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ the horizontal asymptote is y = 0. :::info (4) Find all equations of vertical asymptotes. ::: (4) 1. For the function $f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ the vertical asymptote is x = 0. 2. For the fucntion $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ the vertical asymptote is x = 0. 3. For the function $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ the vertical asymptote is x = 0. :::info (5) Find the interval(s) where $f$ is increasing. ::: (5) 1. $f(x)$ is increasing (-2,0),(0,2) 2.$f'(x)$ is increasing $\left(-\infty,2\sqrt{\left(2\right)}\right),\left(0,2\sqrt{\left(2\right)}\right)$ 3.$f''(x)$ is increasing $\left(\frac{-2\sqrt{\left(30\right)}}{3},0\right),\left(0,\frac{2\sqrt{\left(30\right)}}{3}\right)$ :::info (6) Find the $x$-value(s) of all local maxima. (Find exact values, and not decimal representations) ::: (6) To find the local maxima you would first find where the first derivative is equal to 0. 1. The x-value of the local maxima for the function $f(x)=f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ is 2. 2. There is no local maxima for the function $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$. 3. The x-value for the local maxima for the function $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ is $\frac{2\sqrt{\left(30\right)}}{3}$. :::info (7) Find the $x$-value(s) of all local minima. (Find exact values, and not decimal representations) ::: (7) To find the local minima you would first find where the first derivative is equal to 0. 1. The x-value of the local minima for $f\left(x\right)=\frac{12x^{2}-16}{x^{3}}$ is -2. 2. The local minima for the function $f'\left(x\right)=\frac{12\left(x^{2}-4\right)}{x^{4}}$ is $2\sqrt{\left(2\right)}$. 3. The local minima for the function $f''\left(x\right)=\frac{24\left(x^{2}-8\right)}{x^{5}}$ is $\frac{-2\sqrt{\left(30\right)}}{3}$. :::info (8) Find the interval(s) on which the graph is concave downward. ::: (8) 1. The graph is concave down for the graph $f(x)$ at $\left(-\infty,-2\sqrt{\left(2\right)}\right)and\ \left(0,2\sqrt{\left(2\right)}\right)$ 2. The graph is concave down for the graph $f''(x)$ at $\left(\frac{-2\sqrt{\left(30\right)}}{3},0\right)and\left(0,\frac{2\sqrt{\left(30\right)}}{3}\right)$ 3. The graph is concave down for the graph $f''(x)$ at $\left(-\infty,-2\sqrt{\left(5\right)}\right)and\left(0,2\sqrt{\left(5\right)}\right)$ :::info (9) State the $x$-value(s) of all inflection points. (Find exact values, and not decimal representations) ::: (9) To find the inflection points, you would first find where the second derivative is equal to 0. 1. Inflection point x-value for $f(x)$ is $2\sqrt{\left(2\right)}$ 2. There are no inflection points for $f'(x)$ 3. Inflection point x-value for $f''(x)$ is $2\sqrt{\left(5\right)}$ :::info (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. ::: (10) ![](https://i.imgur.com/TODUtlO.png) --- 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.