rolles theorem

[native geyser]

f(x) = (x - 1) ^ (2/3) - 1 then by rolles theorem
(a) [0, 1]

(b) [0, 2]

(c) [- 1, 1]

(d) none

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[native geyser]

I tried here if we plug in the end points

I am not sure it will be 0 or not

F(0),f(2)

And if we find derivative of f(x)

[cedar tundra]

I assume the question that wasnt posted was "on which interval is rolle's applicable?"

[native geyser]

F'(c)= 2/3((x-1)^(1/3)

[native geyser]

Ohh yes

So if I plug in x=1 in f'(x) we got infinity

[cedar tundra]

yes, f has a cusp at 1

[native geyser]

But we should get a point where f'(c) should be 0

[cedar tundra]

so 1 cant be in the interior of the closed interval, hence b is out (since 1 is in (0,2))

$f(0)=(-1)^{2/3}-1=0\neq f(1)$, so it isnt $[0,1]$

[glacial sparrow]

Omegabet_

[cedar tundra]

and f(-1) clearly isnt f(1), so none of the listed intervals allow rolle's

[native geyser]

Yes true

[cedar tundra]

so d

[native geyser]

Yes that's the answer

[cedar tundra]

yep

[native geyser]

Someone was arguing with me so i posted

Thanks

By the way how did you find it is making an cusp?

I meant without desmos

[cedar tundra]

Handwavy- it's a abs value graph that's been curved

But it doesn't matter it's a cusp, just f isn't differentiable at 1

[native geyser]

@cedar tundra

This one also looks bad question

Because it is not differentiable at x=1

[final relic]

you sure?

yes it's not d-able.

[final relic]

I think for rolle

you need d-ability on (a, b)

and continuity on [a, b]

[final relic]

c, d are not the answer

some of a or b

it's d-able everywhere except x = 1 though.

yeah it's a false problem