Newton's method for approximating zeros is as follows: we select an initial guess for the zero; in this case `x_1=-2` . Then we follow an iterative process to get better and better approximations.

`x_(n+1)=x_n-(f(x_n))/(f'(x_n))`

Here `f(x)=e^x+x^2-4` and `f'(x)=e^x+2x`

(Note that we are trying to find when the graph of the...

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Newton's method for approximating zeros is as follows: we select an initial guess for the zero; in this case `x_1=-2` . Then we follow an iterative process to get better and better approximations.

`x_(n+1)=x_n-(f(x_n))/(f'(x_n))`

Here `f(x)=e^x+x^2-4` and `f'(x)=e^x+2x`

(Note that we are trying to find when the graph of the curve `y=e^x` intersects the parabola `y=4-x^2` . So `e^x=4-x^2 ==> e^x+x^2-4=0` and now we are looking for the zero.)

So `x_1=-2, f(-2)=e^(-2)+(-2)^2-4=1/e^2, f'(-2)=e^(-2)-4=1/e^2-4`

Then `x_2=-2-(1/e^2)/(1/e^2-4)~~-1.9649813665`

The next iteration yields `x_3~~-1.964635631` and my graphing utility yields an approximate answer as -1.964636.

Note that Newton's method does not always work. There are some functions for which the sequence formed by the `x_i's` will not converge. Also there can be dependency on the initial value chosen.

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