 One of the important uses of calculus is in the field of what's called numerical analysis. And roughly speaking, numerical analysis is how you solve an equation that you can't solve algebraically. And we've already seen a little bit of that when we use the intermediate value theorem to find the location of the roots of an equation. But a more powerful method uses the derivative and is sometimes called Newton's method or the Newton-Raphson method. And this begins with the following idea. Suppose f of x is a continuous function where I know that f of 3 is negative 1 and f prime of 3 is equal to 3. Using just this information and a little bit of calculus, we can find an approximate solution to f of x equals 0. And this ties together several important ideas. First, since we know f of 3 is equal to negative 1, we know the graph of y equals f of x passes through the point 3 negative 1. But we're also given the derivative at 3, and the thing to remember is that the derivative is the slope of the line tangent to the curve. Since we know f prime of 3 is equal to 3, we know the line tangent to the graph at this point has slope 3. And so that means we can write the equation of the tangent line. And since the tangent line approximates the curve, we know that the curve looks like not this, not this, but maybe something like this. And the thing to notice is that where the tangent line intersects the x-axis is close to where the graph of the function intersects the x-axis. So an approximate solution to f of x equals 0 can be found by finding where the tangent line intersects the x-axis. So we'll let y equals 0 in the equation of the line and solve for x. And we find x equals 10 thirds, which means that f of x equals 0 has a solution around 10 thirds. Now what's truly impressive about this result is how little we had to know. We had to know the value of the function and the derivative. But beyond that, we knew absolutely nothing about our function f of x. So what if we do know something about f of x? For example, suppose we have our function x cubed minus 7x squared minus 25x plus 61. By using an appropriately chosen tangent line, let's find an approximate solution to g of x equals 0. So let's think about the geometry. If g of x equals 0 and we're graphing y equals g of x, then y is equal to 0. In other words, we're looking for where the curve crosses the x-axis. But since we have calculus, we might remember that the tangent line is a good approximation to the curve through the point of tangency. Since the tangent line approximates the curve near the point of tangency, we'll want to find a tangent line near where the curve crosses the x-axis. And that means we'll want to find a point on the curve that's close to the x-axis. So how can we do that? Well, if you don't play, you can't win. Let's use the intermediate value theorem. We note that g of 0 is 61 and g of, oh, I don't know, how about 5 is equal to negative 114. So let's graph these two points. And since this is a polynomial function, we know it's continuous. And so we know that some place between 0 and 5, the graph crosses the x-axis. Remember that the tangent line is a good approximation to the curve, which means that wherever the tangent line crosses the x-axis, we'll be close to where the curve crosses the x-axis. So let's find where the tangent line crosses the x-axis. So at x equals 0, g of 0 is 61, and we'll need the slope of the tangent line. So we'll differentiate. At x equals 0, we find the derivative is equal to. And so we're able to write the equation of the tangent line. Now the tangent line will intersect the x-axis when y is equal to 0. And so that tells us that x is 61, 25ths. And since the tangent line approximates the curve, that says that x close to 61, 25ths is approximately where g of x equals 0. Now you might wonder how good of an approximation is this. Well, the answer to that is, well, if we can solve this equation and then compare our answers, well, wait a minute. If we could solve this equation, we probably wouldn't be going through this process anyway. So we must assume that we can't actually solve this equation directly, that we're using this as a way to solve the equation. So how can we tell whether we have an approximate solution at g of x equals 0? Well, the simplest way is to evaluate our function at 61, 25ths and see what we get. So we find a value of negative 27 point change. And our solution isn't that good. So you might wonder how we can do better. We'll take a look at that next time.