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

*print*Print*list*Cite

### 2 Answers

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

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

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

=> f'(x) = 48x^3 - 48x

48x^3 - 48x = 0

=> x^3 - x = 0

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

=> x(x - 1)(x + 1) = 0

x = 0

x - 1 = 0 => x = 1

x + 1 = 0 => x = -1

**The solution is x = (-1 , 0 , 1)**

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).**