For what values of x does the equality f’ (x) = 0 hold if it is known that f (x) = 2/x-3/x^2 + 4
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The equation f'(x) = 0 can be solved by taking the derivative of f(x) and setting it equal to 0.
f'(x) = 2/x^2 - 6/x^3
Setting this equal to 0 and solving for x yields two solutions: x = 0 and x = 2.
The value of x = 0 is not valid in this equation, as it would result in division by 0. Therefore, the value of x for which f'(x) = 0 is x = 2.