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OpenStudy (anonymous):

Decide whether the function f=sin(x)+cos(x) satisfies the hypotheses of the MVP on the interval [a,b]=[0,2π] Then find all values of c in the interval [a,b] satisfying f′(c)=(f(b)−f(a))/b−a. If there is more more than one enter them as a comma separated list. c= Enter NONE if there are no such points in the interval.

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

of course is does cosine and sine are both continuous and differentiable for all real numbers, and so certainly continuous aon \([0,2\pi]\)

OpenStudy (anonymous):

you need a couple numbers here \[ f(x)=\sin(x)+\cos(x)\] you need \[f(2\pi)\] and \[f(0)\] what do you get?

OpenStudy (anonymous):

thats the derivative Cos[x] - Sin[x]

OpenStudy (anonymous):

f(2pi) = 1 f(0)= 0 @satellite73

OpenStudy (anonymous):

i don't think so

OpenStudy (anonymous):

\[f(x)=\sin(x)+\cos(x)\] \[f(2\pi)=\sin(2\pi)+\cos(2\pi)\]

OpenStudy (anonymous):

i get \(1\) for that one

OpenStudy (anonymous):

i used the initial equation

OpenStudy (anonymous):

\[f(0)=\sin(0)+\cos(0)

OpenStudy (anonymous):

\[f(0)=\sin(0)+\cos(0)\]

OpenStudy (anonymous):

what did you get for that one?

OpenStudy (anonymous):

lol its 1, my bad

OpenStudy (anonymous):

k good

OpenStudy (anonymous):

your final job is to take the derivative, set it equal to zero and solve

OpenStudy (anonymous):

for 2pi i get 1 also

OpenStudy (anonymous):

right

OpenStudy (anonymous):

Cos[x] - Sin[x]

OpenStudy (anonymous):

and \(1-1=0\) so set the derivative equal to zero

OpenStudy (anonymous):

thats the derivative

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

cos(x) - sin(x) = 0 now what

OpenStudy (anonymous):

i would start with \[\cos(x)=\sin(x)\] and then think of the two points on the unit circle where cosine and sine are the same

OpenStudy (anonymous):

pi/4

OpenStudy (anonymous):

right?

OpenStudy (anonymous):

@satellite73

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