# Why there is no max / min

1. Nov 15, 2009

### jwxie

Given equation
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

2. Nov 15, 2009

### n!kofeyn

When you write an equation like this, use parentheses. I.e. f(x)=3x2/(x-3).

Is this function continuous on the interval [2,8]? If it is, then it must have an absolute maximum and minimum in the interval [2,8], but if not, it may have only an absolute minimum, absolute maximum, or neither.

3. Nov 15, 2009

### jwxie

Thanks for the reminder.
Oh right... 3 is undefined.

Oh I did not notice 3 was also my C.P (dumb)
Thanks!!!

4. Nov 15, 2009

### n!kofeyn

Right! Although, 3 isn't a critical point because it isn't in the domain of the function. Just take the limits as x approaches 3 from the left and right hand sides to show that the function is unbounded in the interval [2,8].