Mean Value Theorem

**Kasper** · July 13th 2009, 03:50 PM

Hey folks, I haven't the foggiest idea how to start this problem. A tip or push in the right direction or link to a resource that explains these types of problems would be just great.

a. Use the Mean Value Theorem to show that for $0<x<y$ ,
$\sqrt{y}-\sqrt{x}<\frac{y-x}{2\sqrt{x}}$ .

b. Use part (a), above, to conclude that for $0<x<y$ ,
$\sqrt{xy}<\frac{y+x}{2}$

I've only done problems with the Mean Value Theorem of the form "Find 2 roots of $f(x)$ and comfirm that $f'(c)=0$ at some point $c$ between those roots.

Even just a hint or link would be great, thanks guys.

**Plato** · July 13th 2009, 04:10 PM

$f(x) = \sqrt x \;\& \,0 < x < y\;\& \,\left( {\exists c \in \left( {x,y} \right)} \right)\left[ {\frac{1}{{2\sqrt c }} = \frac{{\sqrt y - \sqrt x }} {{y - x}}} \right]$

**Kasper** · July 13th 2009, 04:20 PM

Originally Posted by Plato

$f(x) = \sqrt x \;\& \,0 < x < y\;\& \,\left( {\exists c \in \left( {x,y} \right)} \right)\left[ {\frac{1}{{2\sqrt c }} = \frac{{\sqrt y - \sqrt x }} {{y - x}}} \right]$

I'm sorry, I'm lost with this, am I reading this right?

Let $f(x)=\sqrt{x}$ where $0<x<y$ ,

Then there exists a value $c$ in $(x,y)$ where
$ \frac{1}{2\sqrt{c}}=\frac{\sqrt{y}-\sqrt{x}}{y-x}$

But I don't know where that relation came from? I do appreciate the speedy response, that was a real quick answer!

**Krizalid** · July 13th 2009, 05:03 PM

Given $c\in(x,y)$ implies that $\frac{1}{2\sqrt{y}}<\frac{1}{2\sqrt{c}}<\frac{1}{2 \sqrt{x}},$ and you can conclude the inequality from there.

**Kasper** · July 13th 2009, 05:16 PM

Originally Posted by Krizalid

Given $c\in(x,y)$ implies that $\frac{1}{2\sqrt{y}}<\frac{1}{2\sqrt{c}}<\frac{1}{2 \sqrt{x}},$ and you can conclude the inequality from there.

So if I'm understanding this correctly, we are using the mean value theorem (Which states that there is a $c$ in between $x$ and $y$ that makes that inequality valid), to show that
$ f(x)=\sqrt{x}$

$f'(y)<f'(c)<f'(x)$

$\frac{1}{2\sqrt{y}}<\frac{1}{2\sqrt{c}}<\frac{1}{2 \sqrt{x}}$

And then we algebraically manipulate this into the answer?

I'm sorry for the probably extremely basic questions, I'm terrible with the theorems and proofs and need serious work on them.

But also, given the inequality $0<x<y$ in the original equation, shouldn't our derived equality be of the form $f'(x)<f'(c)<f'(y)$ ? How did those values get switched around?

**Krizalid** · July 13th 2009, 05:24 PM

No, actually what I did was by taking $x<c<y\implies 2\sqrt x<2\sqrt c<2\sqrt y,$ and that yields the inequality I gave.

**Kasper** · July 13th 2009, 05:35 PM

Originally Posted by Krizalid

No, actually what I did was by taking $x<c<y\implies 2\sqrt x<2\sqrt c<2\sqrt y,$ and that yields the inequality I gave.

Oh I see, right. Because taking the inverse, the biggest becomes the smallest, the smallest becomes the biggest, etc.

I'm still a little lost as to how we can take the inequality

$ \frac{1}{2\sqrt{y}}<\frac{1}{2\sqrt{c}}<\frac{1}{2 \sqrt{x}} $

to show that

$ \sqrt{y}-\sqrt{x}<\frac{y-x}{2\sqrt{x}} $

for $0<x<y$ .

Sorry, and thanks again for your patience!

**Krizalid** · July 13th 2009, 05:40 PM

Above was stated that $\frac{1}{2\sqrt{c}}=\frac{\sqrt{y}-\sqrt{x}}{y-x}\implies \frac{1}{2\sqrt{y}}<\frac{\sqrt{y}-\sqrt{x}}{y-x}<\frac{1}{2\sqrt{x}}.$

Thus, you can conclude from there.

**Kasper** · July 13th 2009, 05:49 PM

Originally Posted by Krizalid

Above was stated that $\frac{1}{2\sqrt{c}}=\frac{\sqrt{y}-\sqrt{x}}{y-x}\implies \frac{1}{2\sqrt{y}}<\frac{\sqrt{y}-\sqrt{x}}{y-x}<\frac{1}{2\sqrt{x}}.$

Thus, you can conclude from there.

OH! I see!

So the Mean Value Theorem states $f'(c)=\frac{f(b)-f(a)}{b-a}$ where in our case, $b=y$ and $a=x$ and $f(x)=\sqrt{x}$ , thus $f'(c)=\frac{\sqrt{y}-\sqrt{x}}{y-x}$ .

That's where $\frac{1}{2\sqrt{c}}=\frac{\sqrt{y}-\sqrt{x}}{y-x}$ came from, because $f'(c)=\frac{1}{2\sqrt{c}}$

So we substitute this into the inequality you gave me $\frac{1}{2\sqrt{y}}<\frac{\sqrt{y}-\sqrt{x}}{y-x}<\frac{1}{2\sqrt{x}}$ , axe off the very left hand side of the equation and solve for $\sqrt{y}-\sqrt{x}$

which gives us $\sqrt{y}-\sqrt{x}<\frac{y-x}{2\sqrt{x}}$

Which is what we are trying to show is valid under the condition $0<x<y$

Right?

**Krizalid** · July 13th 2009, 05:50 PM

Yes.

**Kasper** · July 13th 2009, 05:53 PM

Originally Posted by Krizalid

Yes.

Awesome! Thanks so much!

**Kasper** · July 13th 2009, 06:05 PM

So real quick sorry, am I correct in this solution to part b?

It says to use part a $\sqrt{y}-\sqrt{x}<\frac{y-x}{2\sqrt{x}} $ , to conclude that for $0<x<y$ , $\sqrt{xy}<\frac{y+x}{2}$ .

So I did;

$\sqrt{y}-\sqrt{x}<\frac{y-x}{2\sqrt{x}} $

$\sqrt{x}\sqrt{y}-x=\frac{y-x}{2}$

$2\sqrt{xy}-2x=y-x$

$2\sqrt{xy}=y+x$

$\sqrt{xy}=\frac{y+x}{2}$

Right?

**Krizalid** · July 13th 2009, 06:17 PM

Yes, it's right but dunno why you turned that $<$ into $=.$

Fixing that your solution is correct.

**Kasper** · July 13th 2009, 06:27 PM

Originally Posted by Krizalid

Yes, it's right but dunno why you turned that $<$ into $=.$

Fixing that your solution is correct.

Oh, sloppy consistency; I'll fix that up in my solution, thanks again for your time and patience, much appreciated.

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