# comepare graph y(x)=X^3-x-1 and tangent lines (x=1, 0.6, 0.57) explain why and how the initial guesses affected the number of iterations needed to converge to the zero of the function based on...

comepare graph y(x)=X^3-x-1 and tangent lines (x=1, 0.6, 0.57)

explain why and how the initial guesses affected the number of iterations needed to converge to the zero of the function based on the tangent line slopes at the initial guess points on the graph

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### 1 Answer

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.