# Solve for x the equation f'(x)=0 if f(x)=14x^4+28x^2-21?

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The function f(x) = 14x^4+28x^2-21.

It is given f'(x) = 0

=> 4*14*x^3 + 28*2*x = 0

=> x^3 + x = 0

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

=> x = 0

x^2 + 1 = 0

=> x^2 = -1

=> x = i and x = -i

**The solution of f'(x) = 0 is x = 0, x = i and x = -i**

To solve the equation f'(x)=0, we need the expression of the 1st derivative.

First, we'll differentiate the function to get the expression of the first derivative.

f'(x) = (14x^4+28x^2-21)'

f'(x) = 14*4*x^3 + 28*2*x - 0

f'(x) = 56x^3 + 56x

We'll impose the constraint from enunciation:

f'(x) = 0

56x^3 + 56x = 0

We'll factorize by 56x:

56x(x^2 + 1) = 0

We'll cancel each factor:

56x = 0

x = 0

x^2 + 1 = 0

x^2 = -1

x = +sqrt(-1)

x = +i

x = -i

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