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Using Newton's Method, we wish to find a solution to:

$$e^x=2-x$$

...starting with $x_0=0$. Find $x_2$.

A

$\dfrac{2-2\sqrt{e}}{2+5\sqrt{e}}$

B

$\dfrac{1}{2} + \dfrac{5-2\sqrt{e}}{2+2\sqrt{e}}$

C

$-\dfrac{1}{2}$

D

$-\dfrac{1}{2} + \dfrac{2-5\sqrt{e}}{2+2\sqrt{e}}$

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