The slope of the tangent line at x is given by `f'(x)=3x^2-1`

An equation of the tangent line is `y=f'(x_0)(x-x_0)+f(x_0)`

At x=1,` y=2(x-1)-1=2x-3`

At x=0.6,` y=0.08(x-0.6)-1.384`

`y=0.08x-1.432`

At `x=0.57, y=-0.0253(x-0.57)-1.38481b`

`y=-0.0253x-1.370386`

The first approximation of Newton's method

gives `x_1=3/2 ` for `x_0=1`

`x_1=1.432/0.08=17.9 ` for `x_0=0.6`

`x_1=-1.370386/0.0253` =`-54.165454` for `x_0=0.57`

Given the graph of the function, the solution of f(x)=0 is between x=1 and x=2.

Starting the process at x_0=1 gives us an iteration x_1=3/2 that approaches the root

whereas for x_0=0.6 or x_0=0.57, the next iteration is further from the solution, thus the number of iterations needed to approach the solution will be bigger.

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