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guys help me find the concavity of f(x)= x^4-4x^3+4x^2. i kow the procedure but have problems with calculations. so helpp

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Differentiate this function twice.
f(x)=x^4-4x^3+4x^2 f'(x)=4x^3-12x^2+8x f''(x)=12x^2-24x+8 let's find the inflection points where the curve changes concavity f''(x)=0 x=1+ 1/root 3 and 1-1/root 3 so x =0.42 and 1.577 here the curve changes concavity if f''(x)>=0 then f(x) is concave up if f''(x)<=0 then f(x) is concave down for the interval (-infinity, 0.42] and [1.577, infinity) f(x)''>=0 for the interval[0.42,1.577] f''(x)<=0 so f(x) has concavity up in the interval(-infinity, 0.42] and [1.577, infinity) and concavity down in [0.42, 1.577]
is it correct to get intervals in decimal??

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yeah, you can also write in radical like 1+1/root 3
plz help me with this f(x)=x^3/(3x^2+1). for the same thing. i messed up while solving.
|dw:1325942133129:dw| First step find the first derivative
|dw:1325942270219:dw| simplify more
Now take the derivative again....
set = to 0 and solve to determine possible points of inflection...then choose values to the left and right of these zeros to determine if x>0 concave up or if x<0 concave down....

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