zeesbrat3 one year ago If f(x) = |(x2 − 4)(x2 + 2)|, how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?

1. zeesbrat3

@Astrophysics @ganeshie8 @Michele_Laino

2. zeesbrat3

help..

3. Michele_Laino

I'm pondering...

4. zeesbrat3

Oh, sorry

5. Michele_Laino

we have to note that we have the subsequent inequality: $\Large {x^2} - 4 < 0,\quad {\text{if }} - 2 < x < 2$

6. Michele_Laino

am I right?

7. zeesbrat3

honestly, im confused

8. Michele_Laino

why?

9. Michele_Laino

it is a simple inequality

10. zeesbrat3

no i understand the inequality, im confused by how you got it

11. Michele_Laino

ok! I wrote that inequality, because inside the interval [0,1] your function can be rewritten as follows: $\Large f\left( x \right) = - \left( {{x^2} - 4} \right)\left( {{x^2} + 2} \right) = \left( {4 - {x^2}} \right)\left( {{x^2} + 2} \right)$

12. zeesbrat3

otherwise any other would be out of bounds

13. Michele_Laino

no, it is a step in order to get rid the absolute value symbol

14. zeesbrat3

oh, okay.

15. Michele_Laino

now, we can apply the theorem af mean value

16. Michele_Laino

so, we have to compute this: $\Large \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}} = \frac{{3 \times 3 - 4 \times 2}}{1} = ...$

17. zeesbrat3

$f'(c) = \frac{ f(b) - f(a) }{ b - a }$

18. zeesbrat3

1

19. Michele_Laino

are you sure?

20. Michele_Laino

$\large \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}} = \frac{{3 \times 3 - 4 \times 2}}{1} = \frac{{6 - 8}}{1} = ...?$

21. zeesbrat3

isnt 3x3 9?

22. Michele_Laino

sorry I have made a big error, you are right

23. zeesbrat3

it's okay :)

24. Michele_Laino

sorry again!!

25. zeesbrat3

no apologies necessary! thank you for your help

26. Michele_Laino

next we have to compute the first derivative at a generic point x, namely we have to compute the first derivative of this function: $\Large f\left( x \right) = \left( {4 - {x^2}} \right)\left( {{x^2} + 2} \right)$

27. Michele_Laino

what do you get?

28. zeesbrat3

$-4x(x^2-1)$

29. Michele_Laino

correct!

30. Michele_Laino

then we have to solve this equation: $\Large - 4x\left( {{x^2} - 1} \right) = 1$

31. Michele_Laino

it is not a simple equation

32. zeesbrat3

$x^3 = -\frac{ 1 }{ 2 }$ ?

33. Michele_Laino

how did you do to write that expression?

34. zeesbrat3

$-4x(x^2-1) = 1$$\frac{ 1 }{ -4x } = x^2$ $\frac{ 1 }{ -4x } +1 = x^2$ $1+1 = -4x^3$ $- \frac{ 1 }{ 2 } = x^3$

35. Michele_Laino

I don't understand this step: $\frac{1}{{ - 4x}} = {x^2}$

36. Michele_Laino

anyway it is not a solution of our equation

37. zeesbrat3

Sorry, i forgot the -1

38. zeesbrat3

How would you get the proper solution?

39. Michele_Laino

I questioned wolfram alpha, here are the solutions:

40. zeesbrat3

So, 3 solutions in the interval

41. Michele_Laino

we have two solutions, inside the interval [0,1]

42. zeesbrat3

Whoops, my bad! I looked outside the interval

43. Michele_Laino

44. Michele_Laino

do you see them?

45. zeesbrat3

Yes. Since one of them is in the negatives, it is not in the interval and does not matter to answer the question

46. Michele_Laino

yes! that's right!

47. zeesbrat3

Thank you!

48. Michele_Laino

:)

49. zeesbrat3

I tried doing the same thing on this problem, but I keep getting no solutions: A particle moves along the x-axis with position function s(t) = xe^x. How many times in the interval [−5, 5] is the velocity equal to 0?

50. Loser66

velocity = (s(t))' take derivative and let it =0 to solve for x .