2006-07-10 08:28:50 · 6 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
y' = 4(x-1)^3
y'' = 12(x-1)^2
y'' = 0 when x = 1
You will see that y'' is never negative. (That is, it doesn't change concavity.)
It will have no inflection point.
2006-07-10 08:33:06 · answer #1 · answered by MsMath 7 · 11⤊ 0⤋
No, it has no inflection point. The second derivative has a zero at x=1, but does not change signs, so x=1 is not an inflection point.
2006-07-10 17:03:01 · answer #2 · answered by mathematician 7 · 0⤊ 0⤋
No. The usual definition of an inflection point is a value x where
1) y(x) exists, AND
2) y'' changes sign at x
For the function in your question, y'' = 12(x-1)^2. Since this is always non-negative, part (2) can never hold and y can't have any points of inflection.
2006-07-10 15:35:32 · answer #3 · answered by Aaron 3 · 0⤊ 0⤋
Take the derivative of the derivative. If there is a zero in there... then yes, there is at least one inflection point.
2006-07-10 15:32:38 · answer #4 · answered by John H 3 · 0⤊ 0⤋
y' = 4(x-1)^3
y'=0 for x = 1
y + 0 +
x 1
y' - 0 +
x 1
y' changes sign from - to + thus x=0 is minimum.
2006-07-10 15:55:19 · answer #5 · answered by gjmb1960 7 · 0⤊ 0⤋
Yes. Of course.
2006-07-10 15:29:43 · answer #6 · answered by aussie_east_ender 2 · 0⤊ 0⤋