Mean Value Theorem - Inequality Problem

**zachd77** · December 27th 2012, 09:25 AM

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!!!

**abender** · December 27th 2012, 05:56 PM

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:

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

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

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

**zachd77** · December 27th 2012, 06:03 PM

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

**Deveno** · December 27th 2012, 11:28 PM

Originally Posted by abender

Now we prove the boundaries:

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

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

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).

Thread: Mean Value Theorem - Inequality Problem

LinkBack

Thread Tools

Display

Mean Value Theorem - Inequality Proof

Re: Mean Value Theorem - Inequality Problem

Re: Mean Value Theorem - Inequality Problem

Re: Mean Value Theorem - Inequality Problem

Similar Math Help Forum Discussions

Proving an inequality using the Mean Value Theorem

Prove inequality using Mean Value Theorem

Proving an inequality using the mean value theorem

using Mean Value Theorem to solve an inequality

Inequality theorem

Search Tags