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solve this to get find global extreme values: f(x)=x^4-8x^2+4x+2 at [-(20/25),(64/25)]
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Why do you specify a point?
is that range?
yep!! extrema values fall in the range so it is specified.
can't solve the cubic is all?
find criticle points check criticle points and boundary points
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i know that but i couldn't solve the equation
D[x^4-8x^2+4x+2 ]-->= 4x^3 -16x+4=0
and how to solve that eqn to get value of x
x^3-4x+1=0...then.. main problem is , to solve for x...
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@sam yes
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