Concavity and Inflection (Medal to the winner. Comments...)
I will give you medal if you show me how to solve this problem, XD
That is why I posted it.
First find first and the second derivative.
After I get the second?
I know f''(x)=-12x^2-36x=0
Once you have the second, find where it's equal to zero, those are your points where it changes concavity
But how do I denote that? X=-3 X=0
Now you have to find where it's concave up and down between those intervals. You could pick random points between them, and find f''
When f'' is positive = concave up f'' negative = concave down
So I choose any number between -3 and 0?
And you want to check outside the interval. The function changes concavity twice.
Alright, so if I chose -2 (f(x)=24) and 3 (f(x)=(-216) is that alright since one is negative and the other is positive?
Still need to check the other side of -3, though. But yes that tells you it's concave up where -3<x<0, and concave down x>0
I'm sorry, how do I check the other side of -3?
The same way you checked the other side of 0 :P pick a number less than -3
I'm sorry xD I'm running out of time with the problem and I don't understand how they want me to denote it in the end.
-3<x<0 is the same as (-3, 0). x>0 means (0, inf)
I never knew that.
How do I get inflection points then?
You already found them lol...
The first things you found!
Oooooooooooookaaaaaaaaaay I seeeeee...so my final answer would be CU: (-3,0) CD: (0,inf) IP: (-3,0)? Sorry if I am getting this wrong, I'm learning this myself. My teacher hasn't been available and his substitute is less than accommodating :S
Never mind, I see it is wrong with the last two ones.
Alright, so I got CU, and IP (they wanted it without the parenthesis)...now to get that CD...thank you for your help thus far by the way...a lot.
Alright, I picked -4 for the CD, and got -48.
CD is two intervals (-inf, -3) as well
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