# User:Saichow2k

## Math 544 exercise

What is the value of f(x) when f(x)=2x+3 at x=0?

## Quiz template

<quiz display=simple> {f(x)=c is a |type="()"} - greatest integer function -identity function +constant function -reciprocal function -absolute value function

In this section we are going to look at a method for approximating solutions to equations. We all know that equations need to be solved on occasion and in fact we’ve solved quite a few equations ourselves to this point. In all the examples we’ve looked at to this point we were able to actually find the solutions, but it’s not always possible to do that exactly and/or do the work by hand. That is where this application comes into play. So, let’s see what this application is all about.

Let’s suppose that we want to approximate the solution to and let’s also suppose that we have somehow found an initial approximation to this solution say, x0. This initial approximation is probably not all that good and so we’d like to find a better approximation. This is easy enough to do. First we will get the tangent line to at x0.

So, how do we find this point?  Well we know it’s coordinates, , and we know that it’s on the tangent line so plug this point into the tangent line and solve for x1 as follows,



{ |type="{}"} The determinant of $\left[{\begin{array}{c c}6&2\\1&4\end{array}}\right]$ is { 22 _3 }.

{Newton's method for solving f(x) = 0 is} +xm+1=xm-f(xm)/f'(xm) -xm+1=xm+f(xm)/f'(xm) -xm+1=xm-f'(xm)/f(xm) -xm+1=xm-f(xm)/f(xm)