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 one year 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)
 one year 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)

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mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so 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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i tried to also find the derivative and i got 1/ (x2)^2

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix please help

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yeah i used the quotient rule

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yeah the first answer was correct.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i 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 . :(

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So 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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0right but wouldn't it be false?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.01=0? doesn't work.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0by undefined what do you mean?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Numbers that would cause a problem :D Like dividing by zero.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Any x values that would cause a problem for our derivative?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Yes! 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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0You stated the domain earlier, so go back and look at it! :)

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0x cannot equal to 2.

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0all real numbers except 2.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So there are no critical points. Should we look for inflection points next? :o

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so theres no relative extrema?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yes last but not least, inflection points. :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So we'll need to find the second derivative for that. Have you tried that step yet?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0no but let me try real quick

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so inflection points is always SECOND DERIVATIVE right?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0y=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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i got (3x^3+17x^232x+20)/(x^24x+4)

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Did you apply the quotient rule? We don't want to do that.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i used the rule book.

zepdrix
 one year ago
Best 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

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

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0don't we plug in severals x's between the intervals or something?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so threres no inflection point?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0between 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Oh the problem didn't ask for concavity did it? _ oh.. hmm

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0include other x's... whu? +_+

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So 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.. :\

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Oh the other curve, umm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0bah i dunno >:O my head hurts.. i need a maf break!
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