Rational functions differentiation (intro)
Worked example: Tangent to the graph of 1/x
In terms of k, where k does not equal 0, what is the y-intercept of the line tangent to the curve f of x is equal to 1/x at the point of the curve where x is equal to k? So let's just think about what they're asking. So if I were to draw myself-- let me draw some quick axes right over here. So that's my y-axis. This is my x-axis right over here. And the graph f of x is equal to 1/x would look something like this. So it looks something like this. So it kind of spikes up there, and then it comes down, and then it goes like this. And I'm just doing a rough approximation of it. So it would look something like that. And then on the negative side, it looks something like this. So this is my hand-drawn version of roughly what this graph looks like. So this right over here is f of x is equal to 1/x. Now, we are concerning ourselves with the point x equals k. So let's say-- I mean, it could be anything that's non-zero, but let's just say that this is k right over here. So that is the point k 1/k. We can visualize the line tangent to the curve there. So it might look something like this. And we need to figure out its y-intercept. Where does it intercept the y-axis? So we need to figure out this point right over here. Well, the best way to do it, if we can figure out the slope of the tangent line here, the slope of the tangent line is just the derivative of the line at that point. If we could figure out the slope of the tangent line, we already know that line contains the point. Let me do this in a different color. We know it contains the point k comma 1/k. So if we know its slope, we know what point it contains, we can figure out what its y-intercept is. So the first step is just, well, what's the slope of the tangent line? Well, to figure out the slope of the tangent line, let's take the derivative. So if we write f of x, instead writing it as 1/x, I'll write it as x to the negative 1 power. That makes it a little bit more obvious that we're about to use the power rule here. So the derivative of f at any point x is going to be equal to-- well, it's going to be the exponent here is negative 1. So negative 1 times x to the-- now we decrement the exponent to the negative 2 power. Or I could say it's negative x to the negative 2. Now, what we care about is the slope when x equals k. So f prime of k is going to be equal to negative k to the negative 2 power. Or another way of thinking about it, this is equal to negative 1 over k squared. So this right over here is the slope of the tangent line at that point. Now, let's just think about what the equation of the tangent line is. And we could think about it in slope-intercept form. So we know the equation of a line in slope-intercept form is y is equal to mx plus b, where m is the slope and b is the y-intercept. So if we can get it in this form, then we know our answer. We know what the y-intercept is going to be. It's going to be b. So let's think about it a little bit. This equation, so we could say y is equal to our m, our slope of the tangent line, when x is equal to k, we just figure out to be this business. It equals this thing right over here. So let me write that in blue. Negative 1 over k squared times x plus b. So how do we solve for b? Well, we know what y is when x is equal to k. And so we can use that to solve for b. We know that y is equal to 1/k when x is equal to k. So this is going to be equal to negative 1-- that's not the same color. Negative 1 over k squared times k plus b. Now, what does this simplify to? See, k over k squared is the same thing as 1/k, so this is going to be negative 1/k. So this part, all of this simplifies to negative 1/k. So how do we solve for b? Well, we could just add 1/k to both sides and we are left with-- if you add 1/k here-- actually, let me just do that. Plus 1/k, left-hand side, you're left with 2/k, and on the right-hand side, you're just left with b, is equal to b. So we're done. The y-intercept of the line tangent to the curve when x equals k is going to be 2/k. If we wanted the equation of the line, well, we've done all the work. Let's write it out. It'll be satisfying. It's going to be y is equal to negative 1/k squared x plus our y-intercept, plus 2/k. And we're done.