Question 1161723
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In the neighborhood of *[tex \Large x\ =\ 2], *[tex \Large f(x)\ =\ -x\ +\ 4] and *[tex \Large g(x)\ =\ x], hence *[tex \Large h(x)\ =\ \frac{-x\ +\ 4}{x}]


So, Low D High minus Hi D Low over Low squared where *[tex \Large 0\ <\ x\ <\ 3]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h'(x)\ = \frac{x(-1)\ -\ (-1\ +\ 4)(1)}{x^2}\ =\ \frac{4}{x^2}\ \forall\ x\ \in\ \mathbb{R}\ |\ 0\ <\ x\ <\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h'(2)\ =\ \frac{4}{2^2}\ =\ 1]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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