- Thread starter
- #1
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 $\dfrac1{\sqrt3} = \dfrac{\sqrt3}3.$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 !