consider the function g(x) = x^3 - (5/2)x^2-2x+1. List any points of inflection. Please show work

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consider the function g(x) = x^3 - (5/2)x^2-2x+1. List any points of inflection. Please show work

Mathematics
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To find the inflection points you have to derive the following function twice. After that, take 2 conditions for f''(x). (a) f''(x) = 0 (b) f''(x) DNE after that take f''(x) = 0 and find the zeros of the function, those zeros will be your inflection numbers, then substitute them in the function to get their y :) and your done ^_^ Give it a try :)
what's DNE
Doesn't exist, and you can use that case if and only if we have x in the denominator :)

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so in your case, you have a function that is considered a polynomial which means the range and domain is R, so you can factor out f''(x) and find the zeros ^_^
clearer? :)
i would factor out x out of the original function but what about the 1
lol, no, to find the inflection points you have to take 2 conditions, like I have said before, but since for x, x exists , you must take the zeros of f''(x) and not f(x) :)
oh you said factor out the second derivative. sry for the misunderstanding
lol, no worries ^_^ yeah, the second derivative = f''(x) :) give it a try now.
ok
good luck :)
thx
np ^_^
the 2nd deriv only gives me one number when = to zero
so that's your inflection number :), I didn't calculate it yet, but if that's what you got, then that's the inflection number ^_^
so i sub that in for all the Xs
yes to get the y, and then you'll have an inflection point (x,y) ^_^
if i substitute it in it will only give me one #, how is that a point? sry if i'm not understanding correctly
no it's alright. Calc dear, when you found the inflection number by getting the zeros, then you have found point x. When you substitute x in the original function, you'll get point y , f(x) = y. Solve for y :)
Did you understand it now? ^_^
gotchya
awesome :)
thx alot that helped
np ^_^ wish you all the best
thnk u
:) np

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