# proof using mean value theorem

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• Nov 15th 2009, 02:24 PM
dgmath
proof using mean value theorem
I need to prove this using the mean value theorem.

1/9 < sqrt(66) - 8 < 1/8

I assumed a = 64 and b = 66 and let a function f(x) = sqrt(x) such that x lies in the interval [a, b]

Then by the mean value theorem,

f'(x) = 1/2sqrt(x) = (sqrt (66) - sqrt (64)) / 66-64 = (sqrt(66) - 8)/2

But at this point, I'm beautifully stuck, and dont know what else to do.

Any ideas?
• Nov 16th 2009, 10:34 AM
Drexel28
Quote:

Originally Posted by dgmath
I need to prove this using the mean value theorem.

1/9 < sqrt(66) - 8 < 1/8

I assumed a = 64 and b = 66 and let a function f(x) = sqrt(x) such that x lies in the interval [a, b]

Then by the mean value theorem,

f'(x) = 1/2sqrt(x) = (sqrt (66) - sqrt (64)) / 66-64 = (sqrt(66) - 8)/2

But at this point, I'm beautifully stuck, and dont know what else to do.

Any ideas?

Problem: Using the MVT prove that $\frac{1}{9}<\sqrt{66}-8<\frac{1}{8}$

Proof: Since $\sqrt{x}$ satisfies all the conditions of the mean value theorem on $[64,66]$ there exists some $c\in(64,66)$ such that $\frac{\sqrt{66}-\sqrt{64}}{64-66}=\frac{1}{2\sqrt{c}}$. Or equivalently, $\sqrt{66}-8=\frac{1}{\sqrt{c}}\quad\color{red}\star$. Since $\frac{1}{\sqrt{x}}$ is a decreasing function on $[64,66]$ we know that for any $k\in(64,66)$ it's true that $\frac{1}{\sqrt{66}}<\frac{1}{\sqrt{k}}<\frac{1}{\s qrt{64}}=8\quad\color{red}\star\star$. Combining $\color{red}\star,\star\star$ we see that $\frac{1}{\sqrt{66}}<\frac{1}{\sqrt{c}}=\sqrt{66}-8<8$. Lastly noting that $\frac{1}{\sqrt{66}}>\frac{1}{\sqrt{81}}=\frac{1}{9 }$ finishes the proof.
• Nov 16th 2009, 04:50 PM
dgmath
oh wow, that makes perfect sense. thanks bunch!