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mathcalculus
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= (5-3x)/(x-2)
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.
i tried to also find the derivative and i got 1/ (x-2)^2
@zepdrix please help
So did you apply the Quotient Rule? One sec, I'll check it :O
yeah i used the quotient rule
Mmm yah I think your first answer was correct: 1/ (x-2)^2
yeah the first answer was correct.
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 . :(
So we have our first derivative:\[\Large y'\quad=\quad \frac{1}{(x-2)^2}\] To find extrema/critical points, we set the first derivative equal to zero and solve for x.
right but wouldn't it be false?
1=0? doesn't work.
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?
by undefined what do you mean?
Numbers that would cause a problem :D Like dividing by zero.
Any x values that would cause a problem for our derivative?
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?
You stated the domain earlier, so go back and look at it! :)
x cannot equal to 2.
Good good. Since it's not in the domain of our function, it can't be a critical point.
all real numbers except 2.
So there are no critical points. Should we look for inflection points next? :o
so theres no relative extrema?
yes last but not least, inflection points. :)
So we'll need to find the second derivative for that. Have you tried that step yet?
no but let me try real quick
so inflection points is always SECOND DERIVATIVE right?
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.
i got (-3x^3+17x^2-32x+20)/(x^2-4x+4)
Dude dude dude -_- what..? Don't expand out squares, especially if they're in the denominator.
Did you apply the quotient rule? We don't want to do that.
i used the rule book.
\[\Large y'\quad=\quad \frac{1}{(x-2)^2}\quad=\quad (x-2)^{-2}\]Power rule from there :O
Ok cool. Setting our second derivative equal to zero:\[\Large 0\quad=\quad \frac{-2}{(x-2)^3}\]
Hmm I think inflection points work the same way as critical points ~ they exist where the second derivative is undefined.. mmmmm
don't we plug in severals x's between the intervals or something?
so threres no inflection point?
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.
Oh the problem didn't ask for concavity did it? -_- oh.. hmm
so help me with inflection point, do we have to include other x's between the intervals?im not sure
include other x's... whu? +_+
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
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.. :\
Oh the other curve, umm
bah i dunno >:O my head hurts.. i need a maf break!