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

1. zzr0ck3r

Is the function continuous on the interval? Is the interval closed?

2. anonymous

um it is continuous and i do believe it is closed

3. zzr0ck3r

Ok, so the theorem applies.

4. anonymous

oh really, okay great that was actually the part i was most unsure of

5. zzr0ck3r

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 $$b=1$$.

6. anonymous

okay so then $\int\limits_{-1}^{1}x^3 -9x$

7. anonymous

so is c the x-coordinate im looking for?

8. zzr0ck3r

$$\int_{-1}^1(t^3-9t) dt = 2(c^3-9c)$$ Correct.

9. anonymous

okay so im getting 2/3 fro F(b) - F(a)

10. anonymous

do i just solve for c now?

11. zzr0ck3r

hmm cubic around the origin shuold even out on that interval

12. zzr0ck3r
13. zzr0ck3r

but yes, solve for $$c$$. I get $$0=2c(c^2-9)$$

14. anonymous

oh woops i see why i got 2/3 , i put t^2 instead of t^3

15. zzr0ck3r

$$-3$$ is not in our interval.

16. anonymous

are you sure i should leave it as zero and not 8.5

17. zzr0ck3r

?

18. anonymous

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 :)

19. anonymous

couldn't c = 0

20. anonymous

$0 = 2c(c^2-9)$ $0 = 2c$ $0 = c$

21. zzr0ck3r

$$0=2c(c^2-9)\\\dfrac{0}{2c}=c^2-9\\0=c^2-9\\c^2=9$$

22. anonymous

c = 3 , 0

23. zzr0ck3r

correct

24. anonymous

so is my answer 0 because 3 is not in the given interval?

25. zzr0ck3r

yes ...sorry

26. anonymous

okay great, thanks for the help :D

27. zzr0ck3r

I think at some point i said $$3\in [-1,1]$$ doooh. deleted it...

28. zzr0ck3r

np m8