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)$(-∞,0)U(0,∞) :::info (2) Find all $x$- and $y$-intercepts. ::: (2) $f(x)=\frac{12x^2-4}{x^4}$ $12(x+2)(x-2)$ $x-2=0$ $x=2 (x-intercept)$ $x+2=0$ $x=-2 (x-intercept)$ $(x^4)=0$ y=0 (no y-intercepts) :::info (3) Find all horizontal asymptotes. ::: (3) $f(x)=\frac{12x^2-16}{x^3}$ Applying the rules of horizontal asymptotes, the horizontal asymptotes is y=0. :::info (4) Find all equations of vertical asymptotes. ::: (4) $f(x)=\frac{12x^2-16}{x^3}$ Applying the rules of vertical asymptotes, the vertical asymptotes is x=0. :::info (5) Find the interval(s) where $f$ is increasing. ::: (5)f is increasing on the interval from (-∞,-2) and increasing (2,∞). ![](https://i.imgur.com/EnTjqLs.jpg) :::info (6) Find the $x$-value(s) of all local maxima. (Find exact values, and not decimal representations) ::: (6)The local maxima is x=-2. ![](https://i.imgur.com/EnTjqLs.jpg) Using the number line to determine where the local maximum is locating when the sign from f' switches from positive to negative. Therefore, x=-2. :::info (7) Find the $x$-value(s) of all local minima. (Find exact values, and not decimal representations) ::: (7)The local minimum is x=2. ![](https://i.imgur.com/EnTjqLs.jpg) Using the number line to determine where the local maximum is locating when the sign from f' switches from negative to positive. Therefore x=2. :::info (8) Find the interval(s) on which the graph is concave downward. ::: (8)The graph concaves downward from $(-∞,-2\sqrt{2})$ and from $(0,2/sqrt{2}). ![](https://i.imgur.com/T9TegHb.jpg) :::info (9) State the $x$-value(s) of all inflection points. (Find exact values, and not decimal representations) ::: (9)The inflection points are when the concavity changes from positive to negative, therefore the inflection points are + and - 2\sqrt{2}. :::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. ::: ![](https://i.imgur.com/EQYm5KD.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.