Prove Inequality with Mean-Value Theorem

[DeusExa]

Homework Statement

Prove that 1/9 ≤ sqrt(66)-8 ≤ 1/8

Homework Equations

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

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!

 

[Johnny Blade]

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]

How did you get a = 64, and b = 66 ?

Do you use sqrtx as function for MVT?

 

[mutton]

sqrt 64 = 8 and sqrt 66 appears in the inequality. Now it should be clear whether sqrt x should be the function.

 

[HallsofIvy]

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?

 

[DeusExa]

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]

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]

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

 

[Johnny Blade]

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]

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]

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

 

[DeusExa]

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]

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]

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?

 

[DeusExa]

bump...have to leave in 10 min

 

[Johnny Blade]

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

 

[DeusExa]

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

=

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

Yay! Proof completed!, right?

 

[Johnny Blade]

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.

 

[DeusExa]

thanks!