A) You need to solve the equation f'(x)=0 to find teh critical values of function such that:

`f'(x) = 4*(2/3)*(x-8)^(2/3-1)`

`f'(x) = (8/3)*(x-8)^(-1/3)`

`f'(x) = 8/(3(x-8)^(1/3))`

`f'(x) = 8/(3root(3)(x-8))`

**Notice that `f'(x)!=0` for any value of x, hence the function has no critical values.**

** B)Notice that the function increases over...**

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A) You need to solve the equation f'(x)=0 to find teh critical values of function such that:

`f'(x) = 4*(2/3)*(x-8)^(2/3-1)`

`f'(x) = (8/3)*(x-8)^(-1/3)`

`f'(x) = 8/(3(x-8)^(1/3))`

`f'(x) = 8/(3root(3)(x-8))`

**Notice that `f'(x)!=0` for any value of x, hence the function has no critical values.**

**B)Notice that the function increases over R set.**

**C)The function does not decrease over R set.**

**D) Since the function has no critical values, hence the function has no maximum or minimum points.**