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anonymous
 one year ago
Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x3 − 9x on the interval [−1, 1]. If so, find the xcoordinates of the point(s) guaranteed to exist by the theorem
anonymous
 one year ago
Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x3 − 9x on the interval [−1, 1]. If so, find the xcoordinates of the point(s) guaranteed to exist by the theorem

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Is the function continuous on the interval? Is the interval closed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0um it is continuous and i do believe it is closed

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Ok, so the theorem applies.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh really, okay great that was actually the part i was most unsure of

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Now we need to look at \[F(b)  F(a)=F'(c)(ba) \] and we have \(F(x) = \int_a^xf(t) dt\) where \(a=1\) and \(b=1\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so then \[ \int\limits_{1}^{1}x^3 9x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is c the xcoordinate im looking for?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(\int_{1}^1(t^39t) dt = 2(c^39c)\) Correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so im getting 2/3 fro F(b)  F(a)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do i just solve for c now?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1hmm cubic around the origin shuold even out on that interval

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=%5Cint_%7B1%7D%5E1%28t%5E39t%29+dt

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1but yes, solve for \(c\). I get \(0=2c(c^29)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh woops i see why i got 2/3 , i put t^2 instead of t^3

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(3\) is not in our interval.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you sure i should leave it as zero and not 8.5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay nvm sorry iv been solving a lot of area and volume problems like these in which area and volume cant be negative so i am told to interpret negatives as positives put i guess it doesn't apply here :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[0 = 2c(c^29)\] \[0 = 2c\] \[0 = c\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(0=2c(c^29)\\\dfrac{0}{2c}=c^29\\0=c^29\\c^2=9\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is my answer 0 because 3 is not in the given interval?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay great, thanks for the help :D

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I think at some point i said \(3\in [1,1]\) doooh. deleted it...
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