Numerical Approximations of Roots
Interactive practice questions
The function $f\left(x\right)=x^3-x^2-3x+15$f(x)=x3−x2−3x+15 has been graphed. Starting with the initial approximation $x_0=-2$x0=−2, we want to use two applications of Newton's method to find a better approximation to the single $x$x-intercept.
Evaluate $f\left(x_0\right)$f(x0)
Evaluate $f'\left(x_0\right)$f′(x0)
Starting with an initial approximation of $x_0=-2$x0=−2, use one application of Newton's method to find a better approximation, $x_1$x1, correct to 2 decimal places if necessary.
Starting with $x_1=-2.69$x1=−2.69, use a second application of Newton's method to find a better approximation, $x_2$x2, correct to 2 decimal places if necessary.
Why would $x_0=0$x0=0 not be a good initial approximation for the $x$x-intercept?
It is on the wrong side of the $x$x-intercept.
Being near a turning point, the next approximation it would generate would be further from the $x$x-intercept rather than closer to it.
The function $f\left(x\right)=3x+7\ln x$f(x)=3x+7lnx has a zero near $x_0=1$x0=1. We want to use one application of Newton's method to find a better approximation to this zero.
The function $f\left(x\right)=2\tan x+\ln x$f(x)=2tanx+lnx has a zero near $x_0=2$x0=2. We would like to use one application of Newton's method to find a better approximation to this zero.
Consider the equation $x^2-13=0$x2−13=0. Starting with the initial approximation $x_0=3.8$x0=3.8, we want to use one application of Newton's method to find a better approximation to the positive root of the equation.