Finding Extrema for (x-1)x^(2/3)

[menal]

Homework Statement

i have to find the extrema for (x-1)x^(2/3)
so i found the derivative: (5x/3)^2/3 - 2/(3x^1/3)
now i have to find the critical numbers by putting the derivative equal to zero and solving for x...but i don't know how to do it.
are there even any roots?

 

[HallsofIvy]

So you have
\left(\frac{5x}{3}\right)^{2/3} - \frac{2}{3x^{1/3}}= 0[/itex]<br /> If x= 0 then the last fraction is undefined (so x= 0 <b>is</b> a critical point), if not we can multiply through by 3x^{1/3}. What does that give you?

 

[menal]

if you plug in zero though, it becomes undefined, not 0...so how can that be a critical point? anddo you mean taking (5x-2)/(3x^1/3)
and since the bottom doesn't matter, jst find the zero of the top part (5x-2) and get 2/5?
would that be correct?
and the only root?

 

[SammyS]

if you plug in zero though, it becomes undefined, not 0...so how can that be a critical point? and do you mean taking (5x-2)/(3x^1/3)
and since the bottom doesn't matter, just find the zero of the top part (5x-2) and get 2/5?
would that be correct?
and the only root?

Review the definition of a critical number when finding extrema.

 

[menal]

oh! lol sorry. i ever knew that! so 0 is a critical number along with 2/5. that would be it?

 

[SammyS]

oh! lol sorry. i ever knew that! so 0 is a critical number along with 2/5. that would be it?

How did you get 2/5?

 

[menal]

"you mean taking (5x-2)/(3x^1/3)
and since the bottom doesn't matter, just find the zero of the top part (5x-2) and get 2/5"

 

[SammyS]

"you mean taking (5x-2)/(3x^1/3)
and since the bottom doesn't matter, just find the zero of the top part (5x-2) and get 2/5"

Right. Now, use the critical numbers to find the extrema.

 

[menal]

yup! the interval given was (0,1)
so i plugged in 0,1 and 2/5 into the first equation:(x-1)x^(2/3)
and got 0,0 and -0.32
so my extreme values would be -0.32 as the absolute minimum and 0 as the absolute max.

 

[SammyS]

Was the interval (0,1), or was it [0,1] ?

Is x = 0 in your interval?

 

[menal]

[0,1]

 

[SammyS]

Look at both ends of the interval.

 

[menal]

you mean -2/5 would be one too?

 

[SammyS]

Of course. But when looking for extrema on a closed interval, you need to check each end of the interval for a possible extremum.

 

[menal]

so your sayin that -2/5 and 2/5 and then 7/5 (1 + 2/5) and 3/5 (1 - 2/5) would all be critical numbers.

 

[SammyS]

No.

If you're looking for extrema on the interval [0,1], the end points of the interval are also critical numbers. The end points are, x=0 (already a critical number) and x=1.

-2/5 is not in the interval [0,1] and not a critical number.

 

[menal]

yes, i have included 0 and 1 and 2/5 as my critical numbers already. I was just confused because i thought u were saying that i had to add/subtract the crticial number i got (2/5) from my intervals (0 and 1). But no, i think i get it now. the crticial numbers will be 0,1 and 2/5.

 

[SammyS]

Yes. 0, 2/5, and 1 .