# Prove Inequality with Mean-Value Theorem

by DeusExa
Tags: inequality, meanvalue, prove, theorem
 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!
 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?
 P: 10 How did you get a = 64, and b = 66 ? Do you use sqrtx as function for MVT?
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.
 PF Patron Sci Advisor Thanks Emeritus P: 38,395 Since $\sqrt{64}= 8$, it should be clear that they are using $f(x)= \sqrt{x}$ on an interval from 64 to 66: (f(66)- f(64))/(66-64)= $(\sqrt{66}- 8)/2$ and that is equal to the derivative, $1/2\sqrt{x}$ for some value of x between 64 and 66. Obviously the largest that can be is $1/2\sqrt{64}= 1/2(8)$. What is the smallest it can be?
 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.
 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.
 P: 10 I don't quite get what you mean, how that applies...please explain...
 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?
 P: 10 So c<66, 66<81, then x<81....then 64
 P: 29 Not really plugging, because you work the MVT with the inequality to prove the initial statement.
 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
 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.
 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
 P: 10 bump...have to leave in 10 min
 P: 29 That's another properties of the inequalities, if 0 < x < y, then 0 < 1/y < 1/x.
 P: 10 then 1/8 < sqrt66-8 < 1/9 = 1/9
 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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