extrema values

[menal]

i was given the equation (x-1)x^(2/3) and told to find the extrema values
and i found the derivative: (5x/3)^(2/3) - (2/(3x^1/3))
and thn i went on to find the critical numbers: 2/5 (5x-2)
and the interval given was: [1,0]
so my critical numbers will be: 2/5, 1, 0
is that it? would 3/5 (1-(2/5)) be included as well?

 

[tkhunny]

and i found the derivative: (5x/3)^(2/3) - (2/(3x^1/3))

Not quite. Get the 5/3 outside the parentheses. This is only a typo, it appears. Pleae be more careful.

and thn i went on to find the critical numbers: 2/5 (5x-2)

I can't decide what this means. x = 2/5 is your suspect? 5x-2 = 0 is what you were solving? Please write things that are maningful and don't require guessing.

and the interval given was: [1,0]
so my critical numbers will be: 2/5, 1, 0

That was quite a leap. What did you do to demonstrate that x = 0 and x = 1 were critical values?

Why are you thinking to include x = 3/5?

 

[tkhunny]

Language Note:

"extrema" is plural

"values" is plural.

"extrema values" sound rather odd.

Stick with either "extrema" or "extreme values".

 

[menal]

1 and 0 are critical numbers because its a given, i did not have to prove this in class so why would i prove it here? my teacher told me that the numbers in the given intervals are included as critical numbers. So thats how i got 1 and 0 from. and the 3/5 i dont know it was a guess. And i added all the parenthesis' to make the equation more clear. and yes i got 2/5 from solving 5x-2=0 for x. so is what i did correct?

 

[tkhunny]

The parentheses around the 5/3 in the expression (5x/3)^(2/3) are incorrect.

3/5 - A guess based on what? Never do that. Knowing is the plan.

Fine, if you are required to include the endpoints, feel free to include them. I don't much liek it, personally, but it serves as a reminde that the calculus doesn't know you limited the Domain. In an assignment searching for extrema, I would consider them "potential critical values". Anyway, I didn't write your textbook.