 So let's see how we can use calculus to improve our solutions to an equation that we can't solve algebraically. And the key to mathematics can be found on a shampoo bottle. Lather, rinse, repeat. Nothing in the rules says we can't repeat this process over and over again. Now one problem we'll run into with any numerical method is the problem of roundoff error. Since we lose accuracy every time we round, we should keep as many decimals as practical. So let's try and keep six decimal places in all of our numbers. Now previously we found the line through x equals sixty one twenty-fifths, that's two point four four zero zero zero, has g of two point four four zero zero, approximately equal to negative twenty-seven point mumble. So we know it passes through this point. But since this is a point on the graph of y equals g of x, we can find the slope of the line tangent to the graph of y equals g of x using the derivative. And so we find the derivative at two point four four zero zero zero is. And we can use this information to find that the line tangent to the graph of y equals g of x at this particular value of x will have equation and the line will intersect the x axis at. And again the reason that this is important is the tangent line approximates the graph. And so that means g of x equals zero will have solution x approximately one point seven eight two six four one. How good a solution is this? Well again we note that if we evaluate our function at one point seven eight two six four one, we find it's approximately equal to where we keep our six decimal places of accuracy. And again it's useful to remember that this is a point on the graph of y equals g of x. And so we can consider the tangent line through this point. So let's try it again. Since the graph of y equals g of x goes through this point, we can find the tangent line using the derivative. And so the derivative at this point is going to be, which allows us to write the equation of the tangent line. And we can determine that this tangent line will intersect the x axis at, which will be an approximate solution to g of x equals zero. How good we'll evaluate g of this number and find it's approximately. So this is an even better approximate solution. And again we note that this also gives us a point on the graph. So let's lather, rinse, repeat. I have a point on the graph. I can find the derivative at that point. I can write the equation of the tangent line. I can find where the tangent line intersects the x axis. And at this point we do run into a slight problem. We can find the same solution as before. And this is because of our round off error. And so because our solution is the same as before due to rounding, we can't get any more accurate with this method. Or put another way, if we do want more accuracy, we needed to have kept more decimal places. We can turn this into a formula. This is known as the Newton-Raphson method. To develop this method, let's consider our steps. The first thing we did was we found an approximate solution where f of x zero was close to zero. Then, since we knew f prime of x, we were able to write the equation of the line through the point x zero, f of x zero, which would have slope f prime of x zero, and equation. And since the line approximates the graph of the function, where the line crosses the x axis, is going to be close to where the graph of the function crosses the x axis. So we found the x intercept. And the important idea here is that this is going to be our next approximate solution. So we can summarize our steps into a formal method. To find increasingly better solutions to f of x equals zero, where f prime of x is continuous, first find an approximate solution x zero, where f of x zero is close to zero. Then, if x n is an approximate solution, then the next approximate solution will be given by this formula. So we can apply our method, but this time let's keep eight decimal places. So our formula requires both the function and the derivative, so we'll go ahead and write those down. We'll start out with x zero equals zero as before, because we knew that zero was an approximate solution. So our first iteration is going to give us x one, approximately. But now I know that two point four four zero zero zero zero zero zero is an approximate solution. So I'll drop that into my formula and get my next approximate solution. But now I know that one point seven eight two six four zero six three is an approximate solution. So I'll drop that into my formula and get the next approximate solution. And I can keep going. But when I get to x five, it's the same as x four, and so I can gain no more accuracy. This is the most accurate I can be with this method. If I want a more accurate solution, I need it to have kept more decimal places.