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f(x) = 3x^2 / x - 3 on [2,8]

I found the critical points were [3, 5/2, 0]

f ' (x) = 3x^2 (x/-5) / (x-3)^2

I want to know why there is no absolute min / max in this case?

I think the reason is because this [a,b] is out of the bound of the critical point, thus we cannot compare [a,b] against the critical points?

Also, is 0 the critical point?

f ' (x) = 3x^2 (x/-5) / (x-3)^2

3x^2 = 0

so i got 0

i always got confused with 1/2 =/ 0

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# Why there is no max / min

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