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

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

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

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

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

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

Now we prove the boundaries:

$\displaystyle 3<c<4 \implies \sqrt{3}<\sqrt{c}<2 \implies \frac{1}{2}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{3}} \implies \frac{1}{4}<\frac{1}{2\sqrt{c}}<\frac{\sqrt{3}}{6}$

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

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

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:

$\displaystyle 3<c<4 \implies \sqrt{3}<\sqrt{c}<2 \implies \frac{1}{2}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{3}} \implies \frac{1}{4}<\frac{1}{2\sqrt{c}}<\frac{\sqrt{3}}{6}$

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

from

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

we have:

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

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:

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

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