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Prove Inequality with Mean-Value Theorem

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that 1/9 ≤ sqrt(66)-8 ≤ 1/8


    2. Relevant equations
    Question comes with:
    Use the mean value theorem
    (f(b) − f(a)) / (b − a)


    3. The attempt at a solution
    I can do "find c" problems fine with MVT, but I have no idea how to prove an inequality. I tried using sqrt(x) as a function, but got stuck. Don't know C.


    Thanks!
     
    Last edited: Dec 2, 2008
  2. jcsd
  3. Dec 2, 2008 #2
    You must remember the interval where c is defined. c is in ]a,b[, where a = 64, and b = 66, now we can write 64 < c < 66. What can you do with the c in this inequality and the one in the MVT?
     
  4. Dec 3, 2008 #3
    How did you get a = 64, and b = 66 ?

    Do you use sqrtx as function for MVT?
     
  5. Dec 3, 2008 #4
    sqrt 64 = 8 and sqrt 66 appears in the inequality. Now it should be clear whether sqrt x should be the function.
     
  6. Dec 3, 2008 #5

    HallsofIvy

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    Science Advisor

    Since [itex]\sqrt{64}= 8[/itex], it should be clear that they are using [itex]f(x)= \sqrt{x}[/itex] on an interval from 64 to 66: (f(66)- f(64))/(66-64)= [itex](\sqrt{66}- 8)/2[/itex] and that is equal to the derivative, [itex]1/2\sqrt{x}[/itex] for some value of x between 64 and 66. Obviously the largest that can be is [itex]1/2\sqrt{64}= 1/2(8)[/itex]. What is the smallest it can be?
     
  7. Dec 3, 2008 #6
    Hi, I used sqrtx and got what you got, being
    1/2sqrtx≤(sqrt(66)-8)/2≤1/2sqrtx

    which then equals

    1/2sqrtx≤(sqrt(66)-8)/2≤1/16

    then, times 2 to clear the 2 in the denominator,

    1/sqrtx≤sqrt(66)-8≤1/8

    This is 2/3 of the proof done, but how do you find the smallest? Logic says that it should be 81, so equals 1/9, which ends the proof, but I'm not sure how.
     
  8. Dec 3, 2008 #7
    Do you remember the properties of the inequalities? The one you need to remember here is if x < y and y < z, therefore x < z.
     
  9. Dec 3, 2008 #8
    I don't quite get what you mean, how that applies...please explain...
     
  10. Dec 3, 2008 #9
    c in the interval ]64, 66[, so we can write 64 < c < 66, but we want 81 in there, 64 < c < 66 < 81. How can you use the property to only have 64, c and 81 in the inequality?
     
  11. Dec 3, 2008 #10
    So c<66, 66<81, then x<81....then 64<c<81, then value of x is between 64 and 81, and then you can plug in 64 and 81 on the two sides of the inequality
    1/2sqrtx≤(sqrt(66)-8)/2≤1/2sqrtx

    giving you 1/9≤sqrt(66)-8≤1/8

    Is that it?
     
  12. Dec 3, 2008 #11
    Not really plugging, because you work the MVT with the inequality to prove the initial statement.
     
  13. Dec 3, 2008 #12
    OK, to reiterate:

    64 < c < 66 (sqrts)

    Then, by using MVT, and 66=a 64=b, you get
    (sqrt66 -8)/2

    Then find derivative of sqrtx, which is 1/2sqrtx.

    Then, using transitive property, prove that
    c<66, 66<81, then x<81....then 64<c<81

    Then,
    1/2sqrtx≤(sqrt(66)-8)/2≤1/2sqrtx

    So, 1/2sqrt81≤(sqrt(66)-8)/2≤1/2sqrt8

    So, times 2 all sides

    Getting 1/sqrt81≤sqrt(66)-8≤1/sqrt64

    Equals 1/9≤sqrt(66)-8≤1/8

    And that is proven. Am I right?
     
  14. Dec 3, 2008 #13
    How did you get this 1/2sqrtx ≤ (sqrt(66)-8)/2 ≤ 1/2sqrtx?

    I would use the MVT to have sqrt(66)-8 = 1/sqrt(c), then transform the inequality 66 < c < 81 to put the sqrt(66)-8 in it.
     
  15. Dec 3, 2008 #14
    My bad.

    64 < c < 66 (sqrts)

    Then, by using MVT, and 66=a 64=b, you get
    (sqrt66 -8)/2

    Then find derivative of sqrtx, which is 1/2sqrtx.

    Then, using transitive property, prove that
    c<66, 66<81, then x<81....then 64<c<81

    then (sqrt66 -8)/2 = 1/2sqrt(c)

    then sqrt(66)-8 = 1/sqrt(c)

    c=1/(sqrt66-8)²

    then 64 < 1/(sqrt66-8)² < 81

    then 1/64 < (sqrt66-8)² < 1/81

    then 1/8 < sqrt66-8 < 1/9

    But that is reversed with what I want to prove
    1/9 ≤ sqrt(66)-8 ≤ 1/8

    ???? What Happened???
     
  16. Dec 3, 2008 #15
    bump...have to leave in 10 min
     
  17. Dec 3, 2008 #16
    That's another properties of the inequalities, if 0 < x < y, then 0 < 1/y < 1/x.
     
  18. Dec 3, 2008 #17
    then 1/8 < sqrt66-8 < 1/9

    =

    1/9<sqrt66-8 < 1/9 !

    Yay! Proof completed!, right?
     
  19. Dec 3, 2008 #18
    It seems right, I would've done it a slightly different way. With 64 < c < 81 I changed it to sqrt(64) < sqrt(c) < sqrt(81), 1/8 > sqrt(c) > 1/9, then with the MVT you have sqrt(66)-8 = 1/sqrt(c), then you substitute 1/8 > sqrt(66) - 8 > 1/9. But that's just me.
     
  20. Dec 3, 2008 #19
    thanks!
     
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