Given f(x)=12x^4+24x^2+56, what is x if f'(x)=0?

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We have the function: f(x)=12x^4+24x^2+56

f(x) = 12x^4 + 24x^2 + 56

f'(x) = 48x^3 + 48x

If f'(x) = 0

=> 48x^3 + 48x = 0

=> x^3 + 3 = 0

=> x( x^2 + 1) = 0

x1 = 0

x2 = -sqrt (-1)

=> x2 = -i

x3 = sqrt (-1

=> x3 = i

Therefore x has the values 0 ,  i , -i

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