# Thread: proof using mean value theorem

1. ## 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?

2. 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 $\displaystyle \frac{1}{9}<\sqrt{66}-8<\frac{1}{8}$

Proof: Since $\displaystyle \sqrt{x}$ satisfies all the conditions of the mean value theorem on $\displaystyle [64,66]$ there exists some $\displaystyle c\in(64,66)$ such that $\displaystyle \frac{\sqrt{66}-\sqrt{64}}{64-66}=\frac{1}{2\sqrt{c}}$. Or equivalently, $\displaystyle \sqrt{66}-8=\frac{1}{\sqrt{c}}\quad\color{red}\star$. Since $\displaystyle \frac{1}{\sqrt{x}}$ is a decreasing function on $\displaystyle [64,66]$ we know that for any $\displaystyle k\in(64,66)$ it's true that $\displaystyle \frac{1}{\sqrt{66}}<\frac{1}{\sqrt{k}}<\frac{1}{\s qrt{64}}=8\quad\color{red}\star\star$. Combining $\displaystyle \color{red}\star,\star\star$ we see that $\displaystyle \frac{1}{\sqrt{66}}<\frac{1}{\sqrt{c}}=\sqrt{66}-8<8$. Lastly noting that $\displaystyle \frac{1}{\sqrt{66}}>\frac{1}{\sqrt{81}}=\frac{1}{9 }$ finishes the proof.

3. oh wow, that makes perfect sense. thanks bunch!

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# prove that 1/9! 1/10! 1/1! = 122/11!

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