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

- anonymous

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

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- jamiebookeater

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- klimenkov

\[\frac{x^2}{(x-5)^2}\]Can you take derivatives?

- anonymous

yes. I know i have to take the second derivative, but i having trouble finding out where the 2nd derivative equals zero.

- anonymous

I even put the function into a graphing calculator. i still cant get the right answer.

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## More answers

- klimenkov

Can you type here what you've got?

- anonymous

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

- klimenkov

Ok. I will type an answer for you.

- anonymous

this was a multiple part question, this is what i have so far.
http://gyazo.com/003eb77ccf9977811aa71cf7f65fb35f

- anonymous

all those answers are right. just need 7 and 8

- klimenkov

\(f''(x)=\frac2{(-5 + x)^2} - \frac{8 x}{(-5 + x)^3} + \frac{6 x^2}{(-5 + x)^4}\)
The horizontal asymtote isn't right.

- klimenkov

Oh. Sorry. I must check it again!

- klimenkov

Sorry, I'm not very good in English. Am I right?|dw:1351890529218:dw|

- anonymous

left is concave downward, right is concave up. so you have it backwards

- anonymous

|dw:1351890663750:dw|

- klimenkov

Ok. So when \(x\in(-\infty,-\frac52)\) it is concave downward. \(x\in(-\frac52,5)\bigcup(5,+\infty)\) it is concave upward.

- klimenkov

I can solve it right here if you want. It will be not so long.

- anonymous

your right. no need to go any further. :P

- klimenkov

All you have to do is to solve this:\[\frac2{(-5 + x)^2} - \frac{8 x}{(-5 + x)^3} + \frac{6 x^2}{(-5 + x)^4}=0\]

- anonymous

ye, i was having a tough time with that. :l

- anonymous

hmm. while your here. could you answer 1 small part to another question? pretty straight forward this time

- klimenkov

Ok, ask anything you like.

- anonymous

http://gyazo.com/3d89eca12903c12ca28e608233e9f102
find critical points

- klimenkov

Hate this congratulations :)

- anonymous

:o?

- anonymous

people told me there are not critical points to that function, but im not entirely sure if their correct.

- klimenkov

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?

- anonymous

3, and -3

- klimenkov

Yes, they are critical.

- anonymous

and 0

- anonymous

but my site says 3, -3 and 0 are not critical points
http://gyazo.com/77d0af7eb8d1e3ce273ff32c24eb58ca

- klimenkov

Are you sure that 0 is a critical point?

- anonymous

your right, its not. f'(0) does not equal 0

- anonymous

but it wont even accept -3 or 3

- anonymous

ugh. wait. i think f'(0)= 0. so it must be a critical point

- klimenkov

No. It is not right. The derivative is:\[\frac{-4 ( x^2+9 )}{( x^2-9 )^2}\]

- anonymous

oh. so what do you do from there. set it equal to zero and solve? that would be -3 and 3 right?

- klimenkov

In points -3 and 3 it doesn't exist, so both of them are critical.

- anonymous

ok so that is the answer?

- klimenkov

Yes.
What is the definition of the critical point on your site?

- anonymous

http://gyazo.com/a531cae84cdcc8b853ca750f1e2b5e0c

- anonymous

im not entirely sure. '_'

- klimenkov

Try to type -3,3 insted of 3,-3

- anonymous

http://gyazo.com/0354e3dc53f2b3172ff9cbbfd156f2a1

- anonymous

this is the problem im having. it wont accept anything.

- klimenkov

Hm..did you type None into this field?

- anonymous

omg. that actually worked

- klimenkov

None?

- anonymous

yes. Thanks a lot for the help =]

- anonymous

i did a combined total of 180 tries on both of the question i asked you. finally got the right answers.

- klimenkov

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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