So at this point of the course, we should be familiar with something known as the secant line, and recall that the secant line is just a line that intersects the curve of our function at two points. Let's say we have this curve down here of the function \( f(x) = x^2 \), we can draw a secant line, which intersects this curve at two specific points. Now in this video, we're going to be learning all about this thing known as the tangent line. While there are some similarities between the secant line and the tangent line, there are also some very key differences, and understanding these differences is going to be very important in this course. So without further ado, let's just jump right into an example to see what these types of problems would look like.
In this example here, we're told given the function \( f(x) = x^2 \), find the slope of the tangent line at \( x = 1 \). Right off the bat, there's already a key difference I noticed with this type of problem. Notice how we are only given one value of interest, one point, \( x = 1 \). For secant lines, we had two values that we were given because there were two points on our graph, but here, we're only given one. And that actually is the key difference between a secant line and a tangent line.
Where a secant line intersects the curve at two points, a tangent line touches it will barely touch the curve at one point. To understand this, let's go ahead and take a look at these differences down here where we can see that we have this secant line. Imagine for the secant line, we take one of these points here, and we just bring it very close to the other point, so it approaches that value, and in essence, we would only be looking at one point on the graph. We can visualize this by taking this secant line and having this point approach the value of \( x = 1 \), having the two points converge very close together.
This is really the main idea of a tangent line. What we could do is only write one point right here at \( x = 1 \), and we could go ahead and draw a line that is tangent to that point. This right here would be the tangent line, and finding the slope of this line would allow us to solve this problem. Now the question, of course, becomes how do we calculate this slope? Well, it turns out it's a very similar process to how we calculated a secant line, except since we only have one point that is approaching the other point, recall this language of approaching we've discussed before when dealing with limits.
So rather than having two points, we're only going to be looking at one point of interest, except we're going to have our entire function approach this limit, where we have the limit as \( x \) approaches this point of interest. So this right here is the equation for finding the slope of the tangent line. Notice how we keep our function and an \( x \) value in there because we're approaching getting very close to that value that we're looking at. Let's go ahead and see if we can calculate the slope of this tangent line. \( c \) is the value of interest, which I can see up here is 1.
What this means is that our \( c \) value is 1. I also see that our function is \( x^2 \). So if I want to find the slope of this tangent line, it is going to be the limit as \( x \) approaches 1 of our function \( f(x) - f(1) \) all divided by \( x - 1 \), and this again is the \( c \) value. Now to do this, I'll go ahead and take our function and plug it in for \( f(x) \). So what that means is we're going to have the function \( x^2 - f(1) \), and \( f(1) \) is just going to be the function evaluated at 1.
We're going to have \( f(1) \), which is going to be \( 1^2 \), and that's simply equal to 1. Now all of this is going to be divided by \( x - 1 \). At this point, what I can do is try applying this limit. But notice if I take this \( x \) approaching 1 here and I plug it in for \( x \), we're going to get \( 1^2 - 1 \) divided by \( 1 - 1 \), which is all just going to become 0 over 0. This is something that we're not allowed to have in mathematics.
So what we need to do is find a way to simplify this and somehow get it so we're not dividing by 0. What I can do is try factoring the top of this fraction, because \( x^2 - 1 \) is the same thing as \( (x-1)(x+1) \). This is the difference of squares, and all of this is going to be divided by \( x - 1 \). Now from here, I can cancel the \( x - 1 \), so all I'm going to have is the limit as \( x \) approaches 1 for \( x + 1 \). Now at this point, I can take this value of 1 and plug it in for \( x \). So we're going to have \( 1+1 \), which is equal to 2. So, 2 is the solution to this problem, and that is the slope of this tangent line.
Notice that we got a slope of 2 when we had the tangent line, but we got a slope of 4 when we had the secant line. That's because we have a steeper slope here since we're looking at two points on the graph, rather than just one point. Now there is one more thing I want to mention, another difference between the secant and tangent line. When calculating the slope of a secant line, we call this the average rate of change. And I think that makes sense since we're, in essence, averaging these two points on the graph.
But if we're looking at the slope of a tangent line, we actually call this the instantaneous rate of change since we're looking at the slope, or the rate of change at just that one value, that one point, so it's instantaneous. Now we also give this a third name, which is called the derivative. So maybe you've heard of derivatives before, but basically, a derivative is just the slope of a tangent line at one point. So this is how you can solve these types of problems, where you need to calculate the derivative, AKA the instantaneous rate of change, AKA the slope of a tangent line. So I hope you found this video helpful, and let's try getting some more practice.
