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Lagrange Theorem

Yankel

Active member
Jan 27, 2012
398
Speaking of theorems, I have another question. I need to show, using Lagrange's theorem, that:

[tex]1.71<\sqrt{3}<1.75[/tex]

By Lagrange's theorem I mean the one of:

f ' (c)=(f(b)-f(a)) / (b-a)

thanks !
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I have never seen the mean-value theorem used in such a way, but I assume we may state:

$\displaystyle f(x)=x^2\,\therefore\,f'(x)=2x$

$\displaystyle a=1.71,\,b=1.75$

Hence:

$\displaystyle c=\frac{1.75^2-1.71^2}{2(1.75-1.71)}=1.73$

$\displaystyle f(c)=1.73^2=2.9929$

Or maybe it's as simple as stating (given the monotonically increasing behavior of the function on the given interval):

$\displaystyle f(1.71)<f(\sqrt{3})<f(1.75)$

$\displaystyle 2.9241<3<3.0625$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Speaking of theorems, I have another question. I need to show, using Lagrange's theorem, that:

[tex]1.71<\sqrt{3}<1.75[/tex]

By Lagrange's theorem I mean the one of:

f ' (c)=(f(b)-f(a)) / (b-a)

thanks !
Use the Lagrange relation (which I prefer to call the mean value theorem) $f'(c) = \dfrac{f(b)-f(a)}{b-a}$, using the function $f(x)=\sqrt x$, and taking $b=4$, $a=3$, so that $c$ has to be some point between 3 and 4. You will then need to estimate the value of $f'(c)$, using the fact that it lies between $f'(3)$ and $f'(4)$ (because $f'(x)$ is a decreasing function in this case). That will give you two inequalities for $\sqrt3$, one of which should lead quite easily to the result $\sqrt{3}<1.75$. The other one is a bit trickier, and you may find it helpful to use the fact that $\dfrac1{\sqrt3} = \dfrac{\sqrt3}3.$
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
it's a darn good thing 12/7 > 1.71 is all i have to say.