 Hi everyone, let's look at a couple of examples of how to use the two limit definitions of a derivative in order to answer some questions. The first one we're going to look at is the curve negative 2x squared plus 3x plus 1. We are asked to find the slope of the tangent to the graph, in other words, the derivative. And then we are asked to determine the location of the horizontal tangent to the curve. So since there is no point at which we are asked to find the specific derivative, it's implied that we're just looking for that general expression that will give us the derivative at any point that we were to substitute in. So that requires us here to use the first limit definition of the derivative. That's the one with the h in it, you will remember. So we're trying to find f prime of x going to be the limit as h approaches zero. Now for the numerator, remember the first part is f of x plus h. So into each of these x's here, you need to substitute x plus h. So let's go to the second slide just so we can write that out for ourselves just so you can see what it looks like. So remember our given expression is negative 2x squared plus 3x plus 1. So when we evaluate that for f of x plus h, we're doing negative 2x plus h the quantity squared plus 3 times x plus h plus 1. So it's really just a matter now of having to multiply all this out. So that binomial squared of course is x squared plus 2x h plus h squared. We can distribute our 3, distribute our negative 2, and that should be about all that we can do. That's a minus. So that becomes the first part of our numerator. So let's go back up and copy that in. And I'm going to put it in parentheses just to offset it a little bit. So it was negative 2x squared minus 4x h minus 2h squared plus 3x. This was from distributing the 3 in the linear term plus 3h plus 1. So that whole thing is f of x plus h. Then we have to do minus the original function. Now I'm going to go ahead and just distribute that negative right away as you do minus this f of x expression. It's going to become a plus 2x squared minus 3x minus 1 and that's all over h. Now if you did your algebra correctly, a lot of terms should cancel out. If they don't, you did something wrong. So the 2x squareds will cancel out here and here. The 3x's will cancel there and there. And then the constants, the ones cancel. So notice all you're left with is a minus 4xh minus 2h squared plus 3h. So let's just rewrite that so you can see pretty clearly what we have. So now you're at the point that you have a limit to evaluate algebraically. Of course, you cannot substitute h0 in for the h because of that denominator. So we are going to have to algebraically manipulate it by factoring out an h from the numerator. And then your h's cancel out. So now you can substitute 0 in place of this h in the numerator. When you do that, you're left with a negative 4x plus 3. And that is our general expression for the derivative of this original function f of x. So what this is giving us is an expression that will help us obtain the derivative, the slope of the tangent to the curve at any x we were to substitute in. Now the follow-up question for this was asking us to determine the location of a horizontal tangent to the curve. We'll think about that. A horizontal line has a slope of 0. This derivative expression here represents the slope of the tangent line to the curve. So if we want the location of a horizontal tangent, hopefully it makes sense that this derivative, this expression that's giving us the slope, has to be equal to 0 because we want it to be a horizontal tangent. So all we need to do is set negative 4x plus 3 equal to 0 and x ends up as three fourths in the end. So it's at three fourths, the x value of three fourths, that we will have a horizontal tangent to the curve. So let's go look at that graphically and we can confirm that is correct. So here I have our f of x function under y equals. So let's take a look at the graph. It's obviously going to be an upside down parabola. So the horizontal tangent to the curve is going to occur at that maximum. That's the only place you will have a horizontal tangent. So let's go ahead and find the maximum and I'll bet you it happens at the x value of three fourths. And we'll just use our calculator features to do that, to find the maximum. So let's see, go a little to the left, a little to the right. And voila, there's your x value of 0.75, three fourths. Let's look at another example. Now this one we are given the specific point at which we want to find the slope of the tangent line. So in this case we can use the second limit definition. So the three is the x value at which we want to find the slope of the tangent. So we're trying to find f prime of three. So remember the way the second limit definition goes. It's the limit as x approaches c. So in this case it's x approaches three of f of x. So that's the original expression minus f of three, f of c. So that's going to be minus a negative five. So of course that becomes plus five over x minus three. Once again you have a limit that you're going to evaluate algebraically. So in the numerator, so we are going to have to factor this. So it factors as x minus three times x minus three. Then that's how the x minus three in the denominator is able to cancel. Of course that's where the quote unquote problem was, because you were not able to substitute three in place of that x in the denominator. So these cancel. So now you can substitute three into the x that remains and notice you get zero. Now think about what that implies. Since we found the derivative to equal zero, think about what kind of line has a slope of zero. We just figured out where it is that this particular parabola has a horizontal tangent line. And it's going to be at that point three comma negative five.