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OpenStudy (anonymous):
For f(x) = cos(x), find all the values of c on the interval (0, pi) that satisfy the Mean Value Theorem.
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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
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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
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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?
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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?
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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?
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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..
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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"
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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
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OpenStudy (anonymous):
okay thanks