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Where is this function concave up and concave down. x^2/(x-5)^2

Mathematics
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\[\frac{x^2}{(x-5)^2}\]Can you take derivatives?
yes. I know i have to take the second derivative, but i having trouble finding out where the 2nd derivative equals zero.
I even put the function into a graphing calculator. i still cant get the right answer.

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Other answers:

Can you type here what you've got?
going to take me a while to get it, have to do chain rule and everything. '_' you can just post the answer if you want. >.> already been on this problem for like an hour
Ok. I will type an answer for you.
this was a multiple part question, this is what i have so far. http://gyazo.com/003eb77ccf9977811aa71cf7f65fb35f
all those answers are right. just need 7 and 8
\(f''(x)=\frac2{(-5 + x)^2} - \frac{8 x}{(-5 + x)^3} + \frac{6 x^2}{(-5 + x)^4}\) The horizontal asymtote isn't right.
Oh. Sorry. I must check it again!
Sorry, I'm not very good in English. Am I right?|dw:1351890529218:dw|
left is concave downward, right is concave up. so you have it backwards
|dw:1351890663750:dw|
Ok. So when \(x\in(-\infty,-\frac52)\) it is concave downward. \(x\in(-\frac52,5)\bigcup(5,+\infty)\) it is concave upward.
I can solve it right here if you want. It will be not so long.
your right. no need to go any further. :P
All you have to do is to solve this:\[\frac2{(-5 + x)^2} - \frac{8 x}{(-5 + x)^3} + \frac{6 x^2}{(-5 + x)^4}=0\]
ye, i was having a tough time with that. :l
hmm. while your here. could you answer 1 small part to another question? pretty straight forward this time
Ok, ask anything you like.
http://gyazo.com/3d89eca12903c12ca28e608233e9f102 find critical points
Hate this congratulations :)
:o?
people told me there are not critical points to that function, but im not entirely sure if their correct.
The definition of the critical point says that it is that point where the derivative of the function is equal to zero or DOES NOT EXIST. It is your case. Wat are the point where it doesn't exist?
3, and -3
Yes, they are critical.
and 0
but my site says 3, -3 and 0 are not critical points http://gyazo.com/77d0af7eb8d1e3ce273ff32c24eb58ca
Are you sure that 0 is a critical point?
your right, its not. f'(0) does not equal 0
but it wont even accept -3 or 3
ugh. wait. i think f'(0)= 0. so it must be a critical point
No. It is not right. The derivative is:\[\frac{-4 ( x^2+9 )}{( x^2-9 )^2}\]
oh. so what do you do from there. set it equal to zero and solve? that would be -3 and 3 right?
In points -3 and 3 it doesn't exist, so both of them are critical.
ok so that is the answer?
Yes. What is the definition of the critical point on your site?
http://gyazo.com/a531cae84cdcc8b853ca750f1e2b5e0c
im not entirely sure. '_'
Try to type -3,3 insted of 3,-3
http://gyazo.com/0354e3dc53f2b3172ff9cbbfd156f2a1
this is the problem im having. it wont accept anything.
Hm..did you type None into this field?
omg. that actually worked
None?
yes. Thanks a lot for the help =]
i did a combined total of 180 tries on both of the question i asked you. finally got the right answers.
It is strange. Ask your teacher the definition of the critical point. Hope you will solve different problems easily in the future. Good luck for you. I wish you a lot of well. Hope you understand.

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