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Mathematics 20 Online
OpenStudy (anonymous):

For f(x) = cos(x), find all the values of c on the interval (0, pi) that satisfy the Mean Value Theorem.

OpenStudy (anonymous):

@electrokid can u help?

OpenStudy (anonymous):

1. find f'(x)

OpenStudy (anonymous):

@onegirl write down the first derivative

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

hold on

OpenStudy (anonymous):

its -sin(x)

OpenStudy (anonymous):

good. now, by MVT, \[f'(x)=\frac{f(b)-f(a)}{b-a}\] where [a,b] = [0,2pi] your domain

OpenStudy (anonymous):

okay so 0 is a and pi b

OpenStudy (anonymous):

so, \[-sin(c)=\frac{\cos(2\pi)-cos(0)}{2\pi-0} \]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so, what is sin(c) ?

OpenStudy (anonymous):

what is it or what equals to it?

OpenStudy (anonymous):

yes. what does it equal to?

OpenStudy (anonymous):

0

OpenStudy (anonymous):

perfect. \[\sin(c)=0\] for what values of "c" in the range of [0,2pi] is sin(c)=0?

OpenStudy (anonymous):

huh? lol sorry i didn't get that last part

OpenStudy (anonymous):

\[\sin(c)=0\] this can happen only for \(c=0,\pi,2\pi,3\pi,...n\pi\) but, we are looking only in the range of [0,2pi] (both EXCLUSIVE)

OpenStudy (anonymous):

ohh ok so 0, pi, and 2pi?

OpenStudy (anonymous):

the interval notations: \[ [a,b]\qquad\text{closed interval includes both a & b}\\ (a,b)\qquad\text{open interval does not include a & b} \]

OpenStudy (anonymous):

what does your question give you? open interval or closed?

OpenStudy (anonymous):

closed

OpenStudy (anonymous):

so, there you go

OpenStudy (anonymous):

you got your three possible values for "c"

OpenStudy (anonymous):

okay but look at these multiply choices for this question. A. c = 0.6901 and 2.4515 B. c = 0 C. c = -2/pi D. c = -0.6901 and -2.4515

OpenStudy (anonymous):

so it will be A?

OpenStudy (anonymous):

we have \(c=\{0,3.14,6.14\}\) from the options, only B has one of our values

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so maybe your interval is something like this: \(0\le c<\pi\) this means, that "0" is included but not "pi"

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

no wait..

OpenStudy (anonymous):

just saw in the question... you domain is \([0,pi)\) not 2pi like we did

OpenStudy (anonymous):

\[\sin(c)=\frac{\cos(\pi)-\cos(0)}{\pi-0}\]

OpenStudy (anonymous):

ohh yes it is!

OpenStudy (anonymous):

so sin(c) is 0

OpenStudy (anonymous):

no \(\cos(\pi)=-1\qquad\cos(0)=1\) \[-\sin(c)=\frac{-1-1}{\pi} \] solve for "c"

OpenStudy (anonymous):

careful with the signs

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

hold on

OpenStudy (anonymous):

I got c = 0.6901 and 2.4515 which will be A

OpenStudy (anonymous):

perfect

OpenStudy (anonymous):

okay thanks

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