- #1
jwxie
- 282
- 0
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
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