Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
PLEASE HELP :( Please help: sketch the graph of a function. label the intercepts, relative extrema, points of inflection, and asymptotes! then state the domain of the function! y= (53x)/(x2)
 5 months ago
 5 months ago
PLEASE HELP :( Please help: sketch the graph of a function. label the intercepts, relative extrema, points of inflection, and asymptotes! then state the domain of the function! y= (53x)/(x2)
 5 months ago
 5 months ago

This Question is Closed

mathcalculusBest ResponseYou've already chosen the best response.0
so i found the vertical asymptote: x=2 and horizontal asymptote: y=3 and the intercepts i found that it is (0, 2.5) and (1.67,0) so the domain is: (inf,2)(2, inf) now I'm stuck on how to do the relative extrema and the points of inflections. :( please help.
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i tried to also find the derivative and i got 1/ (x2)^2
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@zepdrix please help
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
So did you apply the Quotient Rule? One sec, I'll check it :O
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
yeah i used the quotient rule
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Mmm yah I think your first answer was correct: 1/ (x2)^2
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
yeah the first answer was correct.
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i mean I'm trying to find the relative extrema but i forgot how to. i know i have to find the derivative first. then I'm not sure where to go . :(
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
So we have our first derivative:\[\Large y'\quad=\quad \frac{1}{(x2)^2}\] To find extrema/critical points, we set the first derivative equal to zero and solve for x.
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
right but wouldn't it be false?
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
1=0? doesn't work.
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Oh there is another little tid bit that I often forget. critical points also exist when the first derivative is `undefined`. Do you see any points where that might be true?
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
by undefined what do you mean?
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Numbers that would cause a problem :D Like dividing by zero.
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Any x values that would cause a problem for our derivative?
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Yes! Good. x=2 is something we might consider to be a critical point. Before we can say that for certain though, is x=2 in the domain of our function?
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
You stated the domain earlier, so go back and look at it! :)
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
x cannot equal to 2.
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Good good. Since it's not in the domain of our function, it can't be a critical point.
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
all real numbers except 2.
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
So there are no critical points. Should we look for inflection points next? :o
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so theres no relative extrema?
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
yes last but not least, inflection points. :)
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
So we'll need to find the second derivative for that. Have you tried that step yet?
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
no but let me try real quick
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so inflection points is always SECOND DERIVATIVE right?
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
y=0 gives us roots of the function. y'=0 gives us critical points of the function. y''=0 gives us inflection points of the function. yes.
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i got (3x^3+17x^232x+20)/(x^24x+4)
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Dude dude dude _ what..? Don't expand out squares, especially if they're in the denominator.
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Did you apply the quotient rule? We don't want to do that.
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i used the rule book.
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\Large y'\quad=\quad \frac{1}{(x2)^2}\quad=\quad (x2)^{2}\]Power rule from there :O
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Ok cool. Setting our second derivative equal to zero:\[\Large 0\quad=\quad \frac{2}{(x2)^3}\]
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Hmm I think inflection points work the same way as critical points ~ they exist where the second derivative is undefined.. mmmmm
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
don't we plug in severals x's between the intervals or something?
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so threres no inflection point?
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
between what intervals though? We would need to find inflection points to determine which intervals to check. I guess we just check where the function has a discontinuity since we have no inflection points. So ya, check on the left side of 2 and also on the right. We should get different concavity on each side for this problem.
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Oh the problem didn't ask for concavity did it? _ oh.. hmm
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so help me with inflection point, do we have to include other x's between the intervals?im not sure
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
include other x's... whu? +_+
 5 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so this is the answer: it doesn't say which is for which but i got the idea of what is what. my question is how'd they get the other curve on the right side
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
So they labeled the intercepts (as you found), the asymptotes x=2, y=3, and nothing else... So I think we did everything correctly, seeing as how we found no extrema for this one and they didn't label any on their graph.. :\
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Oh the other curve, umm
 5 months ago

zepdrixBest ResponseYou've already chosen the best response.0
bah i dunno >:O my head hurts.. i need a maf break!
 5 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.