Posted by Thomas on Tuesday, May 29, 2012 at 4:08pm.
find the critical points, inflection points, the absolute minimum value of y, and the relative maximum points of y=x^43x^2+2

ap calculus  Reiny, Tuesday, May 29, 2012 at 5:06pm
dy/dx = 4x^3  6x
= 0 for max/mins
2x(2x^2  3) = 0
x = 0 or x = ±√(3/2)
if x = 0, y = 2
if x = +√(3/2) y = ....
if x = √(3/2) y = ....  you do the arithmetic
y'' = 12x^2  6 = 0 for points of inflection
x^2 = 6/12 = 1/2
x = ±√(1/2)
sub back in to get the two points of inflection
I ran it through Wolfram to see the shape of the graph
http://www.wolframalpha.com/input/?i=y%3Dx%5E43x%5E2%2B2
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