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mathcalculus
Group Title
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)
 8 months ago
 8 months ago
mathcalculus Group Title
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)
 8 months ago
 8 months ago

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mathcalculus Group TitleBest 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.
 8 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix please help
 8 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yeah i used the quotient rule
 8 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yeah the first answer was correct.
 8 months ago

mathcalculus Group TitleBest 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 . :(
 8 months ago

zepdrix Group TitleBest 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.
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
right but wouldn't it be false?
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
1=0? doesn't work.
 8 months ago

zepdrix Group TitleBest 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?
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
by undefined what do you mean?
 8 months ago

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
like this one?
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh ya 2!
 8 months ago

zepdrix Group TitleBest 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?
 8 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
x cannot equal to 2.
 8 months ago

zepdrix Group TitleBest 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.
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
all real numbers except 2.
 8 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so theres no relative extrema?
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes last but not least, inflection points. :)
 8 months ago

zepdrix Group TitleBest 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?
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
no but let me try real quick
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so inflection points is always SECOND DERIVATIVE right?
 8 months ago

zepdrix Group TitleBest 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.
 8 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix
 8 months ago

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yeah i did:/
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i used the rule book.
 8 months ago

zepdrix Group TitleBest 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
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ohhhhh yes
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
crap.
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
2/(x2)^3
 8 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ok :D
 8 months ago

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

zepdrix Group TitleBest 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
 8 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so threres no inflection point?
 8 months ago

zepdrix Group TitleBest 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.
 8 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
no it didn't...
 8 months ago

mathcalculus Group TitleBest 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
 8 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
include other x's... whu? +_+
 8 months ago

mathcalculus Group TitleBest 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
 8 months ago

zepdrix Group TitleBest 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.. :\
 8 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Oh the other curve, umm
 8 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
bah i dunno >:O my head hurts.. i need a maf break!
 8 months ago
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