CALCULUS - DIFFERENTIATION 

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find the inflection points f(x) = 3x3-9x2+15x-11
so you take the f ' (x) and then solve for x?
GOOD..YOU ARE PARTLY CORRECT..DONT EQUATE IT TO ZERO AND SOLVE..FIND F " (X) AND EQUATE IT TO ZERO AND SOLVE.
F '(X) = 9X^2-18X+15
F"(X)=18X-18 = 0
X = 1
NOW CONDITIONS FOR INFLECTION POINTS..
1.F"(X)=0 OR DOES NOT EXIST...
HERE IT IS ZERO AT X=1
2.F '(X) AT THE POINT EXISTS..
HERE  F '(1) EXISTS
3.F"(X) SHOULDCHANGE SIGN WHILE PASSING THROUGH THE POINT
HERE POINT IS X=1
F "(X) FOR X<1 = - VE
F"(X) FOR X>1 = +VE
SO IT CHANGES SIGN.
4.THEN THAT POINT IS POINT OF INFLECTION.
SO X=1 AND Y = 3*1^3-9*1^2+15*1-11=3-9+15-11= -2 IS A POINT OF INFLECTION (1,-2)