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 x-coordinates of the point(s) guaranteed to exist by the theorem

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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 x-coordinates of the point(s) guaranteed to exist by the theorem

Mathematics
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Is the function continuous on the interval? Is the interval closed?
um it is continuous and i do believe it is closed
Ok, so the theorem applies.

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oh really, okay great that was actually the part i was most unsure of
Now we need to look at \[F(b) - F(a)=F'(c)(b-a) \] and we have \(F(x) = \int_a^xf(t) dt\) where \(a=-1\) and \(b=1\).
okay so then \[ \int\limits_{-1}^{1}x^3 -9x\]
so is c the x-coordinate im looking for?
\(\int_{-1}^1(t^3-9t) dt = 2(c^3-9c)\) Correct.
okay so im getting 2/3 fro F(b) - F(a)
do i just solve for c now?
hmm cubic around the origin shuold even out on that interval
http://www.wolframalpha.com/input/?i=%5Cint_%7B-1%7D%5E1%28t%5E3-9t%29+dt
but yes, solve for \(c\). I get \(0=2c(c^2-9)\)
oh woops i see why i got 2/3 , i put t^2 instead of t^3
\(-3\) is not in our interval.
are you sure i should leave it as zero and not 8.5
?
okay 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 :)
couldn't c = 0
\[0 = 2c(c^2-9)\] \[0 = 2c\] \[0 = c\]
\(0=2c(c^2-9)\\\dfrac{0}{2c}=c^2-9\\0=c^2-9\\c^2=9\)
c = 3 , 0
correct
so is my answer 0 because 3 is not in the given interval?
yes ...sorry
okay great, thanks for the help :D
I think at some point i said \(3\in [-1,1]\) doooh. deleted it...
np m8

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