## suju101 3 years ago 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

1. across

Differentiate this function twice.

2. ash2326

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]

3. suju101

is it correct to get intervals in decimal??

4. ash2326

yeah, you can also write in radical like 1+1/root 3

5. suju101

plz help me with this f(x)=x^3/(3x^2+1). for the same thing. i messed up while solving.

6. PROSS

|dw:1325942133129:dw| First step find the first derivative

7. PROSS

simplify

8. PROSS

|dw:1325942270219:dw| simplify more

9. PROSS

|dw:1325942373407:dw|

10. PROSS

Now take the derivative again....

11. PROSS

|dw:1325942535869:dw|

12. PROSS

Simplify

13. PROSS

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....