If you explain to me simply, Medal + Fan + Super Testimonial will be given! :) CALCULUS: How do I find the Inflection point? http://screencast.com/t/M74UKjB9B5
@agent0smith @robtobey @ganeshie8 @Mertsj
Inflection point is the second derivative equals to zero.
so how do I actually find the inflection point after I get the seecond derivative? More specifically , what's the inflection point in problem #16 and why?
Set the second derivative equal to zero and solve for x
what if the second derivative doesnt have an x and is f''(x)= -2
There isn't one
The concavity of that function doesn't change
yeah it has a degree of 2, then it is a parabola.
For the function x^2 -3x + 8 that is wrong
This is for problem #16 not #15
The one that is not circled in red
That is also a degree two polynomial and has no inflection point
It indeed says that it does no have an inflection point BUT, it continues and says: F has an inflection point at x = -1/2 whys that
lol nice offer :)
Enters AravindG to save the world once again! :P
if by F they mean the anti-derivative of f, then the inflection point of the anti derivative would be the first derivative of f, which would be -2x -4 = 0 x = -2
@Christos I think Hunus has got hold over the question . I will just watch :)
I dont think integrals are on the game in this problem :/
@AravindG Oki doki :(
I believe that's what they're referring to when they stop using f and say the inflection point of 'F' is -1/2 because f has no inflection points
but the problem itself doesn't ask me to find the inflection point of the anti-derivative
I know. I'm not sure why that is part of their answer
@AravindG do you have something to say/add to this?
can I ADD TO THIS!
Yes of course
because I want medal + FAN + SUPER TESTOMONIAL LOL
okay! lets start with a graph!
ight
go ahead lets see what you got
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