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
find the critical points, inflection points, the absolute minimum value of y, and the relative maximum points of y=x^4-3x^2+2
1 answer