Find f'(x)=0 if f(x)= 12x^4-24x^2+48

To solve the equation f'(x) = 0, we'll have to
differentiate f(x) with respect to x.

f'(x) =
(12x^4-24x^2+48)'

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

dy/dx = 48x^3 - 48x

We'll
cancel f'(x):

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

x1 = +sqrt(1)

x1
=1

x2 = -1

The
real solutions of f'(x) = 0 is: (-1 ; 0 ;
1).