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Prove Inequality with Mean-Value Theorem

by DeusExa
Tags: inequality, meanvalue, prove, theorem
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DeusExa
#1
Dec2-08, 06:14 PM
P: 10
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!
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Johnny Blade
#2
Dec2-08, 08:18 PM
P: 30
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?
DeusExa
#3
Dec3-08, 12:18 AM
P: 10
How did you get a = 64, and b = 66 ?

Do you use sqrtx as function for MVT?

mutton
#4
Dec3-08, 01:23 AM
P: 179
Prove Inequality with Mean-Value Theorem

sqrt 64 = 8 and sqrt 66 appears in the inequality. Now it should be clear whether sqrt x should be the function.
HallsofIvy
#5
Dec3-08, 07:41 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,348
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?
DeusExa
#6
Dec3-08, 06:10 PM
P: 10
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.
Johnny Blade
#7
Dec3-08, 06:33 PM
P: 30
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.
DeusExa
#8
Dec3-08, 06:43 PM
P: 10
I don't quite get what you mean, how that applies...please explain...
Johnny Blade
#9
Dec3-08, 06:50 PM
P: 30
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?
DeusExa
#10
Dec3-08, 06:56 PM
P: 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?
Johnny Blade
#11
Dec3-08, 07:12 PM
P: 30
Not really plugging, because you work the MVT with the inequality to prove the initial statement.
DeusExa
#12
Dec3-08, 07:30 PM
P: 10
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?
Johnny Blade
#13
Dec3-08, 07:47 PM
P: 30
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.
DeusExa
#14
Dec3-08, 08:03 PM
P: 10
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)

then 64 < 1/(sqrt66-8) < 81

then 1/64 < (sqrt66-8) < 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???
DeusExa
#15
Dec3-08, 08:27 PM
P: 10
bump...have to leave in 10 min
Johnny Blade
#16
Dec3-08, 08:31 PM
P: 30
That's another properties of the inequalities, if 0 < x < y, then 0 < 1/y < 1/x.
DeusExa
#17
Dec3-08, 08:34 PM
P: 10
then 1/8 < sqrt66-8 < 1/9

=

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

Yay! Proof completed!, right?
Johnny Blade
#18
Dec3-08, 08:43 PM
P: 30
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.


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