Question

Find the extreme values of the function and where they occur. y=x^3 - 3x^2 +3x-2

Answer 1

Take the derivative; set it to zero; find x.

Answer 2

I did that, and since it came out as a quadratic equation I wanted to make sure I did it right since i got x=1 for both.

Answer 3

you can see the graph at wolfram alpha: http://www.wolframalpha.com/input/?i=y%3Dx%5E3+-+3x%5E2+%2B3x-2

It seems at x=1 there's a point of inflection, and on both sides it increases and decreases for infinity.

Answer 4

the derivative is 3x^2-6x+3
so I got (3x-3)(x-1) therefore x=1 for both

Answer 5

but if i put it as (3x-1)(x-3) wouldn't it be x=1/3 and 3. So I'm just a little confused on which one to use

Answer 6

(3x-1)(x-3) is wrong. (3x)(3)=9x^2

Answer 7

oh oops, so it would be 1. thanks (:

Answer 8

Yeah, but x=1 is a point of inflection, not a maximum or minimum. I'm not entirely sure what the correct answer is. Maybe it is undefined or does not exist?

Answer 9

well i took the 1 and put it in the original equation and got y=-1 so I'm thinking the answer is that the minimum value is -1 at x=1

Answer 10

do you know how to find out if it is a maximum or minimum? Take the second derivative right?

The second derivative is 6x - 6
When x is 1, it is 6(1) - 6 = 0

It is not larger or smaller than zero, so it is not a maximum or minimum. It is a point of inflection. If you follow the wolfram alpha link I gave earlier you can see it.

I think the answer is that there are no global maxima or minima.

Answer 11

oh alright, well thanks for the help

Answer 12

no probs. =)