<p>find area of the region bounded by the graph of f,the x axis, and the lines x=1/k and x=k for k greater than 1
f(x) = x / (1+(x^2))</p>
<p>I'm not sure where you are finding this tough... But here is a solution for you Area - Integration Solution</p>
<p>ooops... remember that its ln (k) for k>1... i forgot to add the k>1 in the final answer.</p>
<p>are u some kind of calculus teacher,tutor etc.. you do solutions right away and have your own website??</p>
<p>Not at all... I graduated in late November, I did tutor throughout Year 12 though...</p>
<p>vrumchev, please help me out on another problem, i would really appreciate your help</p>
<p>Given the function f defined by f(x) = x|x+2| for all x such that -3 <= x <= 1 </p>
<p>a) find the values of x in the given interval for which f is increasing. Justify your answer
b) for what values of x is the graph concave downward?
c) Sketch the graph of f
d) Is f (x) continuous for all x in the given interval? Justify your answer</p>
<p>For absolute value problems, split the function up into a piecewise function</p>
<p>So you would have:</p>
<p>f(x)= x(x+2) -3<=x<=1
x(-x-2) -3<=x<=1</p>
<p>Once you get past that part, the rest of the problem should be fairly simple.</p>
<p>sorry this is going to be quick,</p>
<p>f(x) = -x (x+2) for -3<=x<=-2
= x(x+2) for -2<x<=1</p>
<p>a). f(x) is increasing when f'(x) is positive, ie... -3<=x<2, -1<x<=1
b). -3<=x<=-2
c).....
d). This possible points of discontinuity are the points of 'fracture', ie.. x=-2,-3,1, but we realise that f(x) is continuous at x=-2 as left hand limit = right hand limit = f(-2). Since x=-3 and no left hand limit, and x=1 has no right hand limit, f(x) discontinuous at x=-3,1</p>
<p>plz explain, i understand when f(x) is increasing f ' (x) is positive. So we have x(x+2) AND x(-x-2) for when -3 <= x <= 1
Derivatives are -2x -2 AND 2x+2
so x is greater than -1
so for part a, f(x) is increasing when x is greater than -1 and less than 1. is that correct?</p>
<p>f1'(x) = -2(x+1) is defined for -3<=x<=-2 , therefore, f1'(x) > 0 for -3<=x<=-2...
f2'(x) = 2(x+1) is defined for -2<x<=1, there="" f2'(x)=""> 0 for -1<=x<=1</x<=1,></p>
<p>I made a mistake above and forgot to put in the -ve infront of the 2 for the first restriction in (a).</p>
<p>how did u split the interval into -3<=x<=-2 AND -2<x<=1</p>
<p>x|x+2| is negative for x<-2, and positive for x>-2, and because the questions defines x|x+2| within the domain -3<=x<=1, those restrictions are condensed to -3<=x<=-2 AND -2<x<=1</p>
<p>in part D the question was asking is f (x) continuous for all x in the "given interval", when i graph hte function on my calc the function appears continuous so why did u say discontinuous at x = -3,1</p>
<p>Also please check my answers on this problem, this one is rather easy i hope</p>
<p>given curve x^2 - xy + y^2 = 9</p>
<p>a) find coordinates of points where on the curve the tangents are vertical</p>
<p>i got (y/2 , 2x)</p>
<p>b) at point (0,3) find the rate of change in the slope of the curve with respect to x.</p>
<p>i got 1/2</p>
<p>thanks vrumchev, you have been a great help</p>
<p>Regarding your part (d) in your first question.. remember that for a function to be continuous, both its left hand and right hand limits much exist... Because the question defines f(x) over a domain, the function is not continuous at the extreme points on the domain, ie.. -3 and 1</p>
<p>Question 2).</p>
<p>a). other way around, (2y, x/2).. remember that vertical tangent is dx/dy.. not dy/dx.
b).i'm not quite getting this question.. you're asking me to find the rate of change, and then saying it is with respect to x?</p>
<p>i got question 2, regarding my 1st question part b, u said its concave down -3<=x<=-2 , which isnt possible, calc shows concave down at -2<=x<=-1 i even checked by hand and got the same answer, so did u make a mistake???</p>
<p>also, is it concave down at -2<=x<=-1 OR -2<x<-1 ???</p>
<p>The graph is concave down from -3<=x<=-2</p>
<p>but why is it concave down from -3<=x<=-2
please explain</p>
<p>vrumchev please explain why concave down from -3<=x<=-2<br>
you cant use the 2nd derivative test to figure this out so how did you do it??</p>
<p>Whenever f''(x) < 0 the function is concave down...</p>
<p>For -3<=x<-2, f'(x) = -2x - 2, therefore, f''(x) = -2... ie.. it is concave down.</p>
<p>Alternatively, you can observe the graph (that is how I did it) and in doing so, it becomes clear that the function is concave down for x<=-2....</p>
<p>BTW.. sorry for the late response, but I've been busy the past few days (you can check out my basketball video at my site <a href="http://www.rumchev.com%5B/url%5D">www.rumchev.com</a> :) )</p>