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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2start with \[y=x^2(x2)(x+2)\] so you see that you have 3 zeros, at x =2,x=0,x=2

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2the zero at x = 0 has multiplicity 2, so the graph will touch the x  axis there but not cross it

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2to find the point of inflection you need the second derivative, \[f(x)=x^44x^2\] \[f'(x)=4x^38x\] \[f''(x)=12x^28\] set equal to zero and solve

S
 2 years ago
Best ResponseYou've already chosen the best response.0My main question is that, when i find the inflection point I put square root of 2/3 into the original formula and i get 20/3 right?

S
 2 years ago
Best ResponseYou've already chosen the best response.0And this point can not be on the graph, this makes me confused

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2yes, for the second coordinate you evaluate the function at \[\sqrt{\frac{2}{3}}\]

S
 2 years ago
Best ResponseYou've already chosen the best response.0yes, so I should get 2 inflection points

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2it must be on the graph, plug in the x, find the y

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2right \[\pm\sqrt{\frac{2}{3}}\]

fazna
 2 years ago
Best ResponseYou've already chosen the best response.0for point of inflection we need only to equate first derivative of f(x) to 0.not the second

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2actually it is the second first derivative gives relative max or min

S
 2 years ago
Best ResponseYou've already chosen the best response.0it should be plugged in the original formula, right?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2yes, original function you have the x, you want the y

S
 2 years ago
Best ResponseYou've already chosen the best response.0But in the end my inflection point comes up to be ( square root of 2/3 ;  20/3) and (square root of 2/3 ;  20/3) and my minimum points are (2;4) and (2;4) and it seems that the point of inflection cant be on the graph

fazna
 2 years ago
Best ResponseYou've already chosen the best response.0but im sure that i can get the points of inflection from first derivative,not the second
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