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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giorgiana1976 | Student

First, we'll calculate the first derivative of the given function:

 f'(x) = (12x^4+24x^2+56)'

f'(x) = 12*4*x^3 + 24*2*x + 0

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

We'll impose the constraint from enunciation:

f'(x) = 0

48x^3 + 48x = 0

We'll factorize by 48x:

48x(x^2 + 1) = 0

We'll cancel each factor:

48x = 0

x = 0

x^2 + 1 = 0

x^2 = -1

x = +sqrt(-1)

x = +i

x = -i

The real and complex solutions of f'(x) = 0 are {0 ; +i ; -i}.

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