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If f(x) =4-2x/2 3x 3x^2 then the values of x for which F'(x) =0 is?
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If f(x) =4-2x/2 3x 3x^2 then the values of x for which F'(x) =0 is?
Problem: Find the values of x for which F'(x) = 0 if f(x) = (4-2x)/(2+3x+3x^2).

Solution:

Step 1: Find the derivative of f(x).
F'(x) = [(-2)/(2+3x+3x^2)] - [(4-2x)(3+6x)/(2+3x+3x^2)^2]

Step 2: Set F'(x) = 0 and solve for x.
0 = [(-2)/(2+3x+3x^2)] - [(4-2x)(3+6x)/(2+3x+3x^2)^2]

Step 3: Simplify the equation.
0 = -2(2+3x+3x^2) - (4-2x)(3+6x)
0 = -4 - 6x - 6x^2 - 12 - 24x + 6x + 6x^2
0 = -16 - 30x

Step 4: Solve for x.
x = -16/30
x = -8/15

Step 5: Check for extraneous solutions.
We need to make sure that -8/15 is a valid solution. It is because the denominator of f(x) is never zero, so there are no extraneous solutions.

Conclusion: The value of x for which F'(x) = 0 is -8/15.
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If f(x) =4-2x/2 3x 3x^2 then the values of x for which F'(x) =0 is?
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