# Thread: Mean Value Theorem - Inequality Problem

1. ## Mean Value Theorem - Inequality Proof

Quite an interesting problem:

Use the inequality sqrt(3) < 1.8 to prove that 1.7 < sqrt(3) < 1.75.

The book where I found this says to use the square root function in the MVT for derivatives and use that function on the interval [3,4].

Still confused how to do this.... need help!!!

2. ## Re: Mean Value Theorem - Inequality Problem

Let $f(x)=\sqrt{x}$, $x>0$.

Then, $f^{\prime}(x) = \frac{1}{2\sqrt{x}}$ , $x>0$.

Since $f$ is continuous on $[3,4]$ and differentiable on $(3,4)$, there exists $c\in(3,4)$ such that

$f^{\prime}(c)=\frac{f(4)-f(3)}{4-3}=f(4)-f(3)$.

So, there exists $c\in(3,4)$ such that

$\sqrt{4}-\sqrt{3}=2-\sqrt{3}=\frac{1}{2\sqrt{c}}$.

Now we prove the boundaries:

$3

Since $\frac{1}{2\sqrt{c}}=2-\sqrt{3}$ and the problem gives us that $\sqrt{3}<1.8$ , the last inequality implies

$\tiny{\frac{1}{4}<2-\sqrt{3}<\frac{1.8}{6} \implies 0.25-2<-\sqrt{3}<0.3-2 \implies -1.75<-\sqrt{3}<-1.7 \implies 1.7<\sqrt{3}<1.75}$.

3. ## Re: Mean Value Theorem - Inequality Problem

Thanks!!!! ..... I realized that I did the MVT part wrong.. I solved for c even though I didn't have to...

4. ## Re: Mean Value Theorem - Inequality Problem

Originally Posted by abender

Now we prove the boundaries:

$3

Since $\frac{1}{2\sqrt{c}}=2-\sqrt{3}$.....
note that the information √3 < 1.8 is not needed:

from

$2 - \sqrt{3} < \frac{\sqrt{3}}{6}$

we have:

$12 - 6\sqrt{3} < \sqrt{3} \implies 12 < 7\sqrt{3} \implies \frac{12}{7} < \sqrt{3}$

and 12/7 is approx. 1.7142 which is certainly larger than 1.7 (or 17/10 < 12/7 since: 7*17 = 119 < 120 = 10*12).

we also have:

$\frac{1}{4} < 2 - \sqrt{3} \implies 1 < 8 - 4\sqrt{3} \implies 4\sqrt{3} < 7 \implies \sqrt{3} < \frac{7}{4} = 1.75$

(as a practical matter, this means that 97/56 is a VERY good rational approximation of √3).