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

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}.