Calculus Help?
<div class="IPBDescription">A friend of mine can't figure this out.</div> My friend is in Calculus and being dumb, I can't help. Finding help for the exact type of problem she needs help with online isn't easy to find, and we haven't found anything. This would probably be the best place possible to find help for it. We're not really looking for the answer, but how to solve it.
The problem:
Determine the values of b and c such that the function is continuous on the entire real line.
<!--c1--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>CODE</b> </td></tr><tr><td id='CODE'><!--ec1-->f(x)={ x+1, 1<x<3
x^2+bx+c, |x-2|> or equal to 1<!--c2--></td></tr></table><div class='postcolor'><!--ec2-->
The problem:
Determine the values of b and c such that the function is continuous on the entire real line.
<!--c1--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>CODE</b> </td></tr><tr><td id='CODE'><!--ec1-->f(x)={ x+1, 1<x<3
x^2+bx+c, |x-2|> or equal to 1<!--c2--></td></tr></table><div class='postcolor'><!--ec2-->
Comments
Let's separate this function.
f1(x) = x + 1
f2(x) = x^2 + bx + c
Now, for each of the two points (x=1 and 3), you must ensure that f1 = f2. Also, each of these points' derivates must be equal.
f1(x) = f2(x) and f1'(x) = f2'(x) for x = {1, 3}
Derivating these equations should be easy, and it should yield enough equations for a linear solution to appear.
Intuitively, however, it seems like there are no b and c for which the function is continous, since the derivate of f1 would be 1 for all values of x, and since f2 is a positive parabel, its derivate would always be increasing, thus never equal one more than once.
Let's separate this function.
f1(x) = x + 1
f2(x) = x^2 + bx + c
Now, for each of the two points (x=1 and 3), you must ensure that f1 = f2. Also, each of these points' derivates must be equal.
<!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
I pretty much agree with everything up to this point. However I don't recall continuity requiring that f1'(x) = f2'(x).
So you end up eventually with x^2 -3x +4 for a solution
For a function to be continuous at a point x0 for instance what you are really saying is that : lim x-> x0 f(x) = f(x0). In other words where you expect the function to go by looking at the "run up" to the point x0 is the same as where the function actually ends up i.e. f(x0).
In this case because we are dealing with a function split up into two pieces in order to analyze the "run up" you must look at both functions because you have different definitions for points less than x0 and points greater than x0 . I.E.
lim x->1 f1(x) = f2(x) so this gives us 2 = 1 + b + c (1)
and lim x->3 f1(x) = f2(x) so this gives us 4 = 9 +3b + c (2)
then just solve for b and c given the simultaneous equations (1) and (2)
As far as derivate goes: IMO the only requirement for a continuous function is that a derivative exists at all points. Which it does for these two functions.
Calculus! :O
AHHHHHHHHHHHHHHHHHHHHhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
still have nightmares about math exams
Calculus! :O
AHHHHHHHHHHHHHHHHHHHHhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
Calc? I want to say MATH 124 AHHHHHHHHhhhhhhhhhhhhhhhhhhh newb math!!
That is all.
Let's separate this function.
f1(x) = x + 1
f2(x) = x^2 + bx + c
Now, for each of the two points (x=1 and 3), you must ensure that f1 = f2. Also, each of these points' derivates must be equal.
f1(x) = f2(x) and f1'(x) = f2'(x) for x = {1, 3}
Derivating these equations should be easy, and it should yield enough equations for a linear solution to appear.
Intuitively, however, it seems like there are no b and c for which the function is continous, since the derivate of f1 would be 1 for all values of x, and since f2 is a positive parabel, its derivate would always be increasing, thus never equal one more than once. <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
Dude...dude...
Dude.
No math on the forums.
Ever.
Starting in 4th grade I've always been ahead of my class in Math. I got promoted to 6th grade Maths and was taking basic Algebra and Geometry while in 4th grade. In 5th grade I nearly stumbled when I couldn't memorize my times tables, till I found the pattern for the 9s times tables. In 10th grade i was taking Algebra 2 and Pre Calc
Don't worry, I've taken Calculus and I <i>still</i> don't get it.
Ah, that is correct. My bad -- then a solution absolutely exists.
Don't worry, I've taken Calculus and I <i>still</i> don't get it. <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
I've taken calculus and promptly forgot all about it after passing the last exam.
I win again, education system! You can only strip me of my ignorance for so long!
You do need Calculus to understand what a limit is and how it relates to continuity however. Even though in this case it did not require differentiating/integrating(sp) per se.
Ah, that is correct. My bad -- then a solution absolutely exists. <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
Sounds good <!--emo&:)--><img src='http://www.unknownworlds.com/forums/html/emoticons/smile-fix.gif' border='0' style='vertical-align:middle' alt='smile-fix.gif' /><!--endemo--> Its been a long time since I've done this kinda thing, so I'm glad we reached a consensus.
Don't worry, I've taken Calculus and I <i>still</i> don't get it. <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
I've taken calculus and promptly forgot all about it after passing the last exam.
<b>I win again, education system! You can only strip me of my ignorance for so long!</b> <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
lol that reminds me of the "Laugh it up Vercetti" line.
Ahh blessed ignorance.