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i tried to also find the derivative and i got 1/ (x-2)^2

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.

right but wouldn't it be false?

1=0? doesn't work.

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?

like this one?

oh ya 2!

You stated the domain earlier, so go back and look at it! :)

no.

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?

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.

yeah i did:/

i used the rule book.

\[\Large y'\quad=\quad \frac{1}{(x-2)^2}\quad=\quad (x-2)^{-2}\]Power rule from there :O

ohhhhh yes

crap.

XD

-2/(x-2)^3

ok :D

Ok cool.
Setting our second derivative equal to zero:\[\Large 0\quad=\quad \frac{-2}{(x-2)^3}\]

huh?

don't we plug in severals x's between the intervals or something?

so threres no inflection point?

Oh the problem didn't ask for concavity did it? -_- oh.. hmm

no it didn't...

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? +_+

Oh the other curve, umm

bah i dunno >:O my head hurts.. i need a maf break!