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Homework Help: Mean Value Theorem

  1. Apr 2, 2010 #1
    1. Use the Mean Value Theorem
    f(x) = sqrtX - 2x at [0,4] so a = 0, b = 4

    3. So I found the derivative (which is the slope) and then set the derivative equal to the 1 because of f(b)-f(a)/b-a
    f'(x) = 1/2sqrtX - 2 so 1/2sqrtX - 2 = 0, then 1/2sqrtX = 2, then my solution is x = 1/16

    Just would like a heads up if this is correct, thanks in advanced.
     
  2. jcsd
  3. Apr 2, 2010 #2

    Mark44

    Staff: Mentor

    No, this isn't correct. I think you probably made a mistake in your calculation of (f(b) - f(a))/(b - a), so check your work on this calculation.

    Notice also that the extra parentheses I used (and you didn't) are necessary. The correct interpretation of what you wrote would be
    [tex]f(b) - \frac{f(a)}{b} - a[/tex]

    and I'm sure that's not what you intended.
     
  4. Apr 2, 2010 #3
    well the Mean Value Theorem is 4096de01cffb76de31ec3f510cce7a14.png

    so if a = 0 and b = 4 wouldn't the solution be 1 ? then you set the derivative of the function to 1 which gave me 1/16 ?
     
  5. Apr 2, 2010 #4

    Mark44

    Staff: Mentor

    I think you might be calculating f'(a) and f'(b), instead of f(a) and f(b). Anyway (f(b) - f(a))/(b - a) [itex]\neq[/itex] 1.
     
  6. Apr 2, 2010 #5
    subsitute the values of a and b into the formula
    find the derivative of f(x)
    and after youve got f'(x) and (f(b)-f(a))/b-a
    find your x
     
  7. Apr 2, 2010 #6
    the derivative i got for f(x) is 1/2sqrtX - 2 then i set that equals to (f(b)-f(a))/b-a which is 1, so i'm trying to figure 1/2sqrtX - 2 = 1 and i'm having trouble figuring the fraction
     
  8. Apr 2, 2010 #7
    how are you getting the slope = 1, i got -6/4
    f(4) = 4^1/2 - 8 = -6
    f(0) = 0
    (-6-0)/(4-0) does not equal to 1

    well using my values i get 1/(2(x^1/2)) = -6/4+2
    you can probably work your way from there
     
  9. Apr 2, 2010 #8

    Mark44

    Staff: Mentor

    Stop already! (f(b)-f(a))/(b-a) IS NOT EQUAL TO 1 !

    Put in the actual numbers for a and b and do the calculation!
     
  10. Apr 2, 2010 #9
    You are not taking the derivatives of f(b) and f(a)... the mean value theorem states that if f is continuous on the closed intervals [a,b] which in this case is [0,4] and differentiable on the open interval (a,b,) then there exists a number c in (a,b) such that
    4096de01cffb76de31ec3f510cce7a14.png
     
  11. Apr 2, 2010 #10
    maybe my professor has taught me wrong?
     
  12. Apr 2, 2010 #11
    just look at the formula and substitute a and b... into the equation
     
  13. Apr 2, 2010 #12
    Alright i'm having a brainfart i see what you are saying.
     
  14. Apr 2, 2010 #13
    f(4) = sqrt4 - 8 = -6
    f(0) = sqrt0 - 2(0) = 0
     
  15. Apr 2, 2010 #14
    Alright i think i solved it... finally thanks for the help i really do appreciate it from you 2 :)


    f(4) = -6
    f(0) = 0
    which gives me -6/4

    the derivative 1/2sqrtX -2 = -6/4 which will give me 1/2sqrtx = 1/2 then x = 1
     
  16. Apr 2, 2010 #15

    Mark44

    Staff: Mentor

    Yes.
     
  17. Apr 2, 2010 #16

    Mark44

    Staff: Mentor

    I'm pretty sure he didn't teach you that (2 - 8)/(4 - 0) = 1.
     
  18. Apr 2, 2010 #17
    not so sure about that, sometimes the things they teach you are so obstract that they themselves get confused :confused:
     
  19. Apr 2, 2010 #18

    Mark44

    Staff: Mentor

    No question that an instructor can occasionally get confused with some abstract calculation, but (2 - 8)/(4 - 0) in no way can be considered as abstract.
     
  20. Apr 2, 2010 #19
    indeed it is not abstract, the mean value theorem is very elementary since its dealing with slopes and derivatives so its easy to understand
     
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