Question

f(t)=t^4-12t^3+16t^2

Answer 1

find the critical numbers

Answer 2

oh goody.... the critical numbers are our max min and inflection points.

So off to derivatives we go....

the first derivative tells us where the bends are.

and the second derivative tells us what those bends are doing.

Answer 3

y' = 4t^3 -36t^2 +32t make that equal to 0 to find the bends.

Answer 4

y'' = 12t^2 -72t +32 when this is negative we have a maximum, where it is positive we have a minimum (might have those backwards) and where it is 0 are inflection points

Answer 5

y'' = 4(3t^2 -18t +8)

[18 +-sqrt((18^2) - 4(8)(3))] / 2(3)

3 +-sqrt(324 - 96)/6
3 (+-) 2sqrt(57)/3

2 inflections: x=(3 -2sqrt(57)/3) and (3 +2sqrt(57)/3)

theres the two critical points of inflection... whew this is exhausting :)