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Use the derivative of the function y - f(x) to find the points at which f has a a) local max b) local min c) point of inflection
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y'= (x-1)^2 (x-2)
equate the derivative to 0 to find any max or min equate the second derivative to 0 to find any poi
ohhh ok thanks!@
so it's (x-1)^2 (x-2) = 0 x = 2, 0 and does (x-1)^2 has an x value?
of course it does
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oh so it's -1 and 1 also?
(x-1)^2 = 0 solve for x
(x-1)(x-1)(x-2) = 0 correct? to find max and mins?
since (x-1)^2 is the same as (x-1) (x-1)
^ no (x^2 -1) = (x-1)(x+1)
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how is it x^2 - 1? if the squared is outside the parenthesis?? it's not (x^2-1) because it says (x-1)^2? isn't that different? :0
what i'm saying is (x-1)^2 =/= (x-1)(x+1)
does not equal?
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