Consider the roots (zeros) of f(x)=13x3−4x+1f(x)=13x3-4x+1. 1234-1-2-3-41234567-1-2-3-4-5 We will see that small changes
Posted: Thu Jun 30, 2022 7:41 pm
Consider the roots (zeros) of f(x)=13x3−4x+1f(x)=13x3-4x+1. 1234-1-2-3-41234567-1-2-3-4-5 We will see that small changes in the choice of xoxo produce different roots, or none at all. For each xoxo given, state the root (zero) of f(x)f(x) to which the algorithm converges, or write DNE if it does not converge. round to 3 decimal places If xo=1.85xo=1.85, then Newton's Method converges to: x=x= If xo=1.7xo=1.7, then Newton's Method converges to: x=x= If xo=1.55xo=1.55, then Newton's Method converges to: x=x=
Consider the roots (zeros) of f(x)=13x3−4x+1f(x)=13x3-4x+1. 1234-1-2-3-41234567-1-2-3-4-5 We will see that small changes in the choice of xoxo produce different roots, or none at all. For each xoxo given, state the root (zero) of f(x)f(x) to which the algorithm converges, or write DNE if it does not converge. round to 3 decimal places If xo=1.85xo=1.85, then Newton's Method converges to: x=x= If xo=1.7xo=1.7, then Newton's Method converges to: x=x= If xo=1.55xo=1.55, then Newton's Method converges to: x=x=
Consider the roots (zeros) of f(x)=13x3−4x+1f(x)=13x3-4x+1. 1234-1-2-3-41234567-1-2-3-4-5 We will see that small changes in the choice of xoxo produce different roots, or none at all. For each xoxo given, state the root (zero) of f(x)f(x) to which the algorithm converges, or write DNE if it does not converge. round to 3 decimal places If xo=1.85xo=1.85, then Newton's Method converges to: x=x= If xo=1.7xo=1.7, then Newton's Method converges to: x=x= If xo=1.55xo=1.55, then Newton's Method converges to: x=x=