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Determine the critical points of the following function on \[R\] \[f(x)=x^{4}+2x^{3}-2x^{2}+1\]
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\[f'(x)=2x+6x^2-4x=6x^2-2x=0---->\]
can u solve for x now
i am sorrry i wrote wrong..i edit the question just a minute pls
now its ok...
do it the sameway differentiate and equate to 0 and solve for x
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\[f'(x) = 4x^{3}+6x^{2}-4x\] \[f''(x) = 12x^{2}+12x-4\] and now ?
\[4x^3+6x^2-4x=0..x(4x^2+6x-4)=0......solve\]
there is in solution following numbers \[x_{1}=-\frac{-3-\sqrt{15}}{4} \] \[x_{2}=0\] \[x_{3}=\frac{-3+\sqrt{15}}{4}\] but i dont understand how its possible to find this solutions...
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