Describe the x-values at which the function is differentiable: (Enter your answer using interval notation) f(x) = (3x)^(2/3) (-∞, 0) U (0, ∞)
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Using the chain rule, we get: f'(x) = (2/3)((x) 3)-1/3 * 3 Simplifying, we get: f'(x) = 2(x) -1/3 Now, we need to find the x-values at which this derivative exists. The only way the derivative wouldn't exist is if the denominator of the fraction is 0. So, we Show more…
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