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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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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