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Q: Help With calculus1 Problem \$30.00 ( Answered ,   1 Comment )
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 Subject: Help With calculus1 Problem \$30.00 Category: Science > Math Asked by: alwin86-ga List Price: \$30.00 Posted: 11 May 2006 16:33 PDT Expires: 10 Jun 2006 16:33 PDT Question ID: 727902
 ```Hello, I need help with my calc 1 Problem. Analyze and sketch the graph of the function f (x) = (2x-1)/x Analyze meaning finding the 1. Domain 2. Locate x and y intercepts 3. Check for symmetry 4. Find all horizontal and vertical asymptotes 5. Locate all critical numbers 6. Find all intervals on which f is increasing or decreasing. 7. Find all local max and min 8. Find all intervals on which f is concave up and concave down. 9. Find all points of inflection``` Clarification of Question by alwin86-ga on 11 May 2006 16:38 PDT ```I would like not only the answers but the procedure of how I should get them as well, Thanks...``` Clarification of Question by alwin86-ga on 11 May 2006 17:19 PDT `Please somebody help me, I need this before tomorrow friday at 10:00 AM.`
 ```Hi!! Thank you for asking to Google Answers. Below you will find the requested answer. 1. Domain: Domain is the set of values of the independent variable for which the function is defined. In this case the numerator is defined for all the real numbers, but the division is not defined when the denominator is zero, so this function is not defined when its denominator (x) is equal to zero. Therefore, the domain of f(x) = (2x-1)/x is the set of the real numbers except the zero. 2. Locate x and y intercepts: A x-intercept is a value of x for which the function is zero, so we must solve for f(x)=0 : (2x-1)/x = 0 <==> 2x-1 = 0 <==> 2x = 1 <==> x = 1/2 1/2 is the only x-intercept. An y-intercept is the value of the function when x is zero, but remember that f(x) is not defined at such point, so you do not have any y-intercept for this function. 3. Check for symmetry: See the following definition: "DEFINITION. A function f is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if f(-x) = f(x) for all x in the domain of f. A function f is odd if the graph of f is symmetric with respect to the origin. Algebraically, f is odd if and only if f(-x) = -f(x) for all x in the domain of f." From "Visual Calculus - Even and Odd Functions": http://archives.math.utk.edu/visual.calculus/0/functions.14/index.html Since f(1/2) = 0 and f(-1/2) = 4, f is neither an even nor an odd function. You can also see: "Symmetry of graphs - Topics in precalculus": http://www.themathpage.com/aPreCalc/symmetry.htm 4. Find all horizontal and vertical asymptotes: x = c is a vertical asymptote if the following are all true: 1) f(c) is undefined 2) limit when x tends to c- of f(x) is +? or -? 3) limit when x tends to c+ of f(x) is +? or -? The only choice here is x=0, the only point where f is undefined: 1) f(0) is undefined 2) limit when x tends to 0- of f(x) is +? 3) limit when x tends to c+ of f(x) is -? We have a vertical asymptote at x=0. If the limit when x tends to +? of f(x) = L OR the limit when x tends to -? of f(x) = L, then the line y = L is a horizontal asymptote of f. limit when x tends to +? of f(x) = 2 limit when x tends to -? of f(x) = 2 The line y = 2 is a horizontal asymptote of f. See for references this page: "1.1.5.1 Limits At Infinity And Infinite Limits": http://www.geocities.com/pkving4math2tor1/1_lim_and_cont/1_01_05_01_lim_at_inf_and_inf_lim.htm 5. Locate all critical numbers: Given a real number t. If f'(t) = 0 or f' is undefined at t, then t is a critical number of f. So start calculating f'(x): f'(x) = [(2x-1)/x]' = (2 - 1/x)' = (2)' - (1/x)' = 1/x^2 Note that there is no number for which f'(x) = 0 and f'(x) is only undefined at x=0, then x=0 as the only critical number of f (at a later point you will see that for x<0 and x>0 the convexity of the function changes). 6. Find all intervals on which f is increasing or decreasing: Suppose that for a given interval is: 1) the first derivative of f is positive, then f is increasing; 2) the first derivative of f is negative, then f is decreasing. f'(x) = 1/x^2 > zero in (-?,0) and (0,+?) (where f is defined), then f is increasing in in all its domain. 7. Find all local max and min: Since we have no critical numbers, except the zero when f is not defined, we have no local max and min. Note that because f is increasing in in all its domain, for a given closed interval [a,b](a and b different to zero), the minimum is f(a) and the maximum is f(b). 8. Find all intervals on which f is concave up and concave down: Suppose that for a given interval is: 1) the second derivative of f is positive, then f is concave up; 2) the second derivative of f is negative, then f is concave down. f''(x) = -2/x^3 , then: f''(x) > 0 <==> x < 0, i.e. in (-?,0) and f''(x) < 0 <==> x > 0, i.e. in (0,+?) So in (-?,0) f is concave up and in (0,+?) f is concave down. 9. Find all points of inflection: A point of inflection is a point at which the graph changes from being concave up to concave down, or vice versa. You can use the second derivative to find them following this simple rule: If (c, f(c)) is a point of inflection on the graph of f, then either f''(c) = 0 or is undefined at x = c. Note that at a point of inflection the second derivative must change its sign (this is the test). f''(x) = -2/x^3 , it is never zero and it is only undefined at x=0, so zero is POSSIBLY a point of inflection. But f is not defined at zero so it cannot be a point of inflection, then f has no inflections points. For references see the following pages: "Graphing Using First and Second Derivatives": A page with link to great solved examples!!! http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html "SparkNotes: Applications of the Derivative (AB)": Start a navigation here to find explanations and examples related to this problem. http://www.sparknotes.com/math/calcab/applicationsofthederivative/index.html Search strategy: I used the following keywords at Google.com: function second derivative function first derivative points of inflection intercepts x y critical numbers horizontal vertical asymptotes domain function I hope this helps you. Feel free to use the clarification feature if you find something wrong, unclear or incomplete. I will be glad to give you further assistance on this question if you need it. Best regards, livioflores-ga function symmetry``` Clarification of Answer by livioflores-ga on 11 May 2006 23:29 PDT ```Hi again!! This is a clarification about the answer given for the point 5) Locate all critical numbers and how to graph the function. See this definition: If f is defined at c and either f'(c)=0 or f'(c) is undefined, then c is a critical number of f. This means that a critical number is a number ON THE DOMAIN of a function where the derivative is equal to zero or it is undefined. x=0 is not on the domain on the function f, so it cannot be a critical number according to this definition. So for this problem f has no critical numbers. Some definitions does not include the "ON THE DOMAIN of f" condition and lead me to identify zero as a critical number of f. But in mathematics the correct definition must include that condition, and a critical number must be on the domain of the function. Regarding the graph of f(x) I cannot include one here because the answer box formatting prevents it. To see how the function's graph is go to the following page and input the following string: (2*x-1)/x ; 2 The first part is the function f and the second part will represent the horizontal asymptote. Lewt the other number in the box as they appear by default and click on the "Graph it" button. You must have a Java enabled browser. You can enlarge the graph's window by drap and dragging its borders. "Function Grapher": http://people.hofstra.edu/staff/steven_r_costenoble/Graf/Graf.html You can also try this one, you can only input (2*x-1)/x and click the Draw button; use the Zoom out button if you cannot see the graph clearly. Then input 2 and click the Draw button again. "maths online function plotter": http://www.univie.ac.at/future.media/moe/fplotter/fplotter.html I hope this clarifies the answer. Regards, livioflores-ga```
 alwin86-ga rated this answer: ```Good Job I would like you to get my e - m- a- i- l a j m g 8 6 at m s n dot c o m so we could later on thanksss.```
 ```Looks like homework. 1. Domain. Where is the function defined? Not where you divide by zero. 2. The x intercept is the value of x for which y is zero. This is easily seen by inspection. For the y intercept, consider what you learned in 1. 3. The asmyptotes will in 4. will help you find the symmetry. 4. The horizontal asymptote will be found in the value of y as x approaches infinity. The vertical asymptote will be found as the value of the denominator approaches zero. 5. Critical numbers are maximums, minimums, points of inflection. Use first and second deriviatives. The remaining questions are really contained in question 5.```