Question 1166990
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ e^x\ -\ x\ -\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(1)\ =\ e\ -\ 2\ \approx\ 0.718281]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(0)\ =\ 1\ -\ 0\ -\ 1\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{df}{dx}\ =\ e^x\ -\ 1\ \geq\ 0\ \text{on}\ 0\ \leq\ x\ \leq\ 1] 


Therefore the function is constantly increasing on the given interval, hence *[tex \Large x\ =\ 0] is the absolute minimum on the given interval and is therefore the only zero of the function on the interval.


So why do you need Newton-Raphson?

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish]


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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