Solution:

Given: f(x) = 4-2x/ 2 3x 3x²

We are asked to find the values of x for which f'(x) = 0.

Step 1: Find f'(x)

To find f'(x), we need to use the quotient rule of differentiation. The quotient rule states that if f(x) = u(x)/v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)]/v^2(x).

Applying the quotient rule to f(x), we get:

f'(x) = [(2)(3x)(3x²) - (-2)(2(3x) + (4-2x)(6x))]/(2(3x)³)

Simplifying this expression, we get:

f'(x) = (36x³ + 24x² - 36x² + 24x)/(18x³)

f'(x) = (12x + 8)/(6x²)

Step 2: Find the values of x for which f'(x) = 0

To find the values of x for which f'(x) = 0, we need to solve the equation (12x + 8)/(6x²) = 0.

We can simplify this equation by multiplying both sides by 6x²:

12x + 8 = 0

Subtracting 8 from both sides, we get:

12x = -8

Dividing both sides by 12, we get:

x = -2/3

Therefore, the only value of x for which f'(x) = 0 is x = -2/3.

Conclusion: The only value of x for which f'(x) = 0 is x = -2/3.

Let us take X value as 2 ,
f(2)=4-2(2)/2 *3(2) *3(2)^ ,
=4-4/2*6*12 ,
=0/144 ,
f(2)= 0 . so 'X' value is 2 .