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^4-3x^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%5E4-3x%5E2%2B2

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