
#1
Dec208, 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! 



#2
Dec208, 08:18 PM

P: 29

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?




#3
Dec308, 12:18 AM

P: 10

How did you get a = 64, and b = 66 ?
Do you use sqrtx as function for MVT? 



#4
Dec308, 01:23 AM

P: 179

Prove Inequality with MeanValue Theorem
sqrt 64 = 8 and sqrt 66 appears in the inequality. Now it should be clear whether sqrt x should be the function.




#5
Dec308, 07:41 AM

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PF Gold
P: 38,879

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))/(6664)= [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?




#6
Dec308, 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. 



#7
Dec308, 06:33 PM

P: 29

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.




#8
Dec308, 06:43 PM

P: 10

I don't quite get what you mean, how that applies...please explain...




#9
Dec308, 06:50 PM

P: 29

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?




#10
Dec308, 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? 



#11
Dec308, 07:12 PM

P: 29

Not really plugging, because you work the MVT with the inequality to prove the initial statement.




#12
Dec308, 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? 



#13
Dec308, 07:47 PM

P: 29

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. 



#14
Dec308, 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/(sqrt668)² then 64 < 1/(sqrt668)² < 81 then 1/64 < (sqrt668)² < 1/81 then 1/8 < sqrt668 < 1/9 But that is reversed with what I want to prove 1/9 ≤ sqrt(66)8 ≤ 1/8 ???? What Happened??? 



#15
Dec308, 08:27 PM

P: 10

bump...have to leave in 10 min




#16
Dec308, 08:31 PM

P: 29

That's another properties of the inequalities, if 0 < x < y, then 0 < 1/y < 1/x.




#17
Dec308, 08:34 PM

P: 10

then 1/8 < sqrt668 < 1/9
= 1/9<sqrt668 < 1/9 ! Yay! Proof completed!, right? 



#18
Dec308, 08:43 PM

P: 29

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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