At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
@Mertsj can u help?
@experimentX how would i do this one?
let f(x) = sin(x) then f'(x) = ?
when is cos(x) zero?
zero? you said find the derivative of sinx right?
yes ... at what point is cos(x) equal to zero?
there is not point where it is equal to zero
when i graph it...it doesn't
you got the wrong graph then.
|dw:1363197320465:dw| thats how it looks liek ohh okay
yea it is
okay i got it
so ... what's the point?
so its more than 1?
so its 1.57080?
so pick any two points above pi/2 and below pi/2
okay? do you do this different than the las tone tho? i'm not gonna use f(a) - f(b)/ b - a ?
|dw:1363198546707:dw| do you get that?
yes i do
Okay i gotta go ...
So i filled in those a b c's and here i got....f(pi/2) = f(3pi/4) - f(pi/4) over 3pi/4 - pi/4
f(3pi/4) = 3pi^2 + 1 = ?
can u tell me if i plugged it right @experimentX ?
yes that's correct.
\[ f(3 \pi /4) = \sin(3\pi/4) = 1/ \sqrt 2 \] Find the value of f(pi/4) from your calculator or wolfram.
I got 1
huh .. do it again, what is sin(pi/4) ... plug into wolfram.
i got 1 over sqrt of 2
what do i do next ?
yes ... plug into your Rolle's condition.
the formula right?
so f(1/sqrt(2) - f(1/sqrt(2)/ 1/sqrt(2) - 1/sqrt(2) ?
how do i solve that?
so it will be 3/3 right?
so those will be the numerator?
yes ... you just found out the value. Do you know what is functional notation?
yeah ... that is the problem. you haven't advanced to the point of doing this. let f(x) = cos(x), what is value of f(0)
f(x) = 1
yep ... if f(x) = sin(x) ,,, then what is f(pi/4) ??
1 over sqrt(2)
yes ... what is f(3pi/4) ??
its the same 1 over sqrt(2)
yes ... plug in those values in |dw:1363385976116:dw|
plug them in the denominator?
ohh okay hold on
f(1/sqrt(2) - f(1/sqrt(2) over 3pi/4 - pi/4
No no ... what are you dong?? why are you putting 1/sqrt(2) inside function?
you said to plug them in?
yes .. |dw:1363386528821:dw|
i'll ask my teacher on this one thanks for your help.
so i plug those in pi?
no ... you just calculated those values. plug them in.
I just did and you told me it was incorrect..i'm confused
yes ... this is where you got confused.
what is the value of|dw:1363388256685:dw|
yes ... so |dw:1363388362361:dw|
the derivative of it?
yes ... is there a point where derivative is zero?
what is that point?
umm at 0, 0?
we talked about that in the above posts. check again ... the point is pi/2
ohh yes okay
so it reached 0 at pi/2
no ... the derivative is zero at pi/2
gotta go ... you gotta work harder.
ok well can u tell me what i should do next before you go
you wanted to verify Rolle's theorem, which you did.
yea its continuos and differentiable i know that
what do you want to do?
find a value for c
so this could be true a < c < b
of course it is ..
ok so c is pi/2 ?
so c is 2pi/4
see you around ... gotta go offline :)