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) I found the domian by setting the denominator equal to zero. $x^3=0$ so x=o. This means that x=0 the function is undefined. Domain[x<0 or x>0 ] :::info (2) Find all $x$- and $y$-intercepts. ::: (2) There is no y intercept because the function will be undefined when you set x=0. The x intercept is found when you set the function equal to zero. $0=\frac{12x^2-16}{x^3}$ I set the numerator equal to zero, $12x^2-16=0$ $12x^2=16$ $x^2=16/12$ $x=-√(4/3)$ and $x=√(4/3)$ are the x intercepts. :::info (3) Find all equations of horizontal asymptotes. ::: (3) The equation of the horizontal asymptote is y=0 I found this because the horizontal asymptote is y=0 is when the numerators degree is one less than the degree of the numerator. :::info (4) Find all equations of vertical asymptotes. ::: (4) The equation of the verticle asymptote is x=0, because this is when the function is undefined. :::info (5) Find the interval(s) where $f$ is increasing. ::: (5) I found the intervals by finding the critical values of the function by settinf f'(x)= 0 or undefined f is increasing from -2<x<0 and 0<x<2 :::info (6) Find the $x$-value(s) of all local maxima. (Find exact values, and not decimal representations) ::: (6) The local maxima is (2,4) I got this by inputting the extreme point back into the function. $y=12(2)^2-16/(2)^3$ :::info (7) Find the $x$-value(s) of all local minima. (Find exact values, and not decimal representations) ::: (7) The local minimum x values is -2, I got this by utilizing the other extreme point. the coordinate is (-2,-4) :::info (8) Find the interval(s) on which the graph is concave downward. ::: (8) I first found the inflection points then plugged in to see if it was positive or negative. The graph is concave down in the interval (-infinty<x< -2√2) and (0<x<2√2) :::info (9) State the $x$-value(s) of all inflection points. (Find exact values, and not decimal representations) ::: (9) I found the inflection points by making the f"=0 and finding where f" is undefined by making the denominator equal to zero. The inflection points are at x=-2√2 and x=2√2, Then I plugged those into to find the y coordinate. Points are (-2√2, -5/√2) and (2√2, 5/√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. ::: (10) <iframe src="https://www.desmos.com/calculator/ggplofn0i9?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe> --- 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.