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You need to evaluate the critical values, hence, in order to do so, you must solve for x the equation `f'(x) = 0` .
You need to find `f'(x), ` such that:
`f'(x) = (1/3)(x-3)^(1/3-1) => f'(x) = 1/(3*(root(3)(x-3)^2))`
You need to evaluate `f'(x) = 0` :
`1/(3*(root(3)(x-3)^2)) = 0`
There is no value of x to verify the equation `f'(x) = 0` , hence, there are no critical points.
Notice that the derivative is positive on R, hence, the function is increasing on R.
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