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

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

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

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

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

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

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0by undefined what do you mean?

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

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

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0all real numbers except 2.

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

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0no but let me try real quick

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

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

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

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

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

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

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

mathcalculus
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0include other x's... whu? +_+

mathcalculus
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Oh the other curve, umm

zepdrix
 2 years 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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