## anonymous one year ago Graph the function f(x)=x^3−2x and it's secant line through the points (-2,-4) and (2,4). Use the graph to estimate the x-coordinate of the points where the tangent line is parallel to the secant line. Find the exact value of the numbers c that satisfy the conclusion of the mean value theorem for the interval [-2,2].

1. anonymous

I need help finding 'c'

2. anonymous

I know that the Mean Value Theorem says... $\frac{ f(b)-f(a) }{ b-a }=f'(c)$

3. anonymous

just apply that...

4. anonymous

But that gives me f'(c)

5. anonymous

lemme check

6. anonymous

here a=-2 b=2

7. anonymous

Right, plugging all that in do you get f'(c) = 2 ?

8. anonymous

i get 8

9. anonymous

How?

10. anonymous

f(b)=2^3-2(2)..rite?

11. anonymous

f(a)=(-2)^3-2(-2)

12. anonymous

Doesn't it already give us the coordinates for the points? (2,4) and (-2,-4)

13. anonymous

Is it wrong to say $\frac{ 4+4 }{ 2+2}=\frac{ 8 }{ 4 } = 2$ ? Or am I totally wrong right now? Lol

14. anonymous

what did u do just now?

15. myininaya

$(x^3-2x)'|_{x=c}=\frac{f(2)-f(-2)}{2-(-2)} \\ (x^3-2x)'|_{x=c}=\frac{4-(-4)}{2-(-2)} \\ (x^3-2x)'|_{x=c}=\frac{4+4}{2+2}$ that is right for the right hand side

16. anonymous

Well I plugged in f(a) = -4, f(b) = 4, and a=-2, b=2 in the mean value theorem formula

17. anonymous

what happens to the left hand side?

18. anonymous

do i just derive it?

19. myininaya

you still have to differentiate the left hand side and plug in c afterwards as the equation above suggests

20. anonymous

3x^2 - 2 right?

21. myininaya

yep and plug in c (though that part doesn't matter to much; it is just what they call it in the formula above) yep now solve 3x^2-2=2

22. myininaya

or you named it c 3c^2-2=2

23. anonymous

Ohh so then its +/- 2/sqrt3 right?

24. myininaya

are both of those in the given interval?

25. anonymous

hmmm... yes?

26. myininaya

ok then cool stuff

27. anonymous

Thats it? :O

28. myininaya

that is it

29. myininaya

though there are two problems

30. myininaya

did you do the first problem?

31. anonymous

Yay!!! Thats correct! :D Thank you so much!

32. myininaya

we did the second problem we found the exact solutions

33. anonymous

Yeah Im always better in graphing haha

34. myininaya

alright

35. anonymous

Thanks again for the help! :)

36. myininaya

np

37. anonymous

tracy?got the answer for the second part?

38. anonymous

Yeah! I got it, its + or - 2/sqrt3

39. anonymous

Thank you too @chethus!