 So let's put this together and see if we can form a graph from the derivative and other information. So one thing that we might want to incorporate is where the derivative fails to exist. And so remember that the derivative can fail to exist even if the function exists, if the function is not continuous, or if the tangent line is vertical, or if the slopes of the secant line disagree. Nevertheless, we can still construct the graph of the function from its derivatives. For example, suppose I have information that looks something like this. Let's try and sketch a graph that incorporates all of the given information. So to begin with, we know there's a discontinuity of some sort at x equals 3, so we'll mark it. Because our graph is discontinuous at x equals 3, we'll want to indicate that we cannot cross this line at x equals 3. So maybe we'll do this as a red dashed line as a reminder. Do not cross this line. What else do we know? At x equals 0, we know the value of the first derivative and the second derivative. So we know that the first derivative is positive, and so we know that the tangent line slopes upward, and we know the second derivative is positive, and so we know the tangent line is below the curve. And so we can sketch a portion of the curve that runs through x equals 0. Now remember that while we can put a point here and run the curve through it, the vertical position of this point needs to be adjusted at the end of the problem. At x equals 2, we know the first derivative is positive, and the second derivative is negative, and so we know the tangent line slopes upward, and the tangent line is above the curve. And so we can sketch the curve through x equals 2. Finally, at x equals 4, we know the first derivative is negative, and the second derivative is negative, and so we know, and we can sketch the curve through x equals 4. And finally, we have this peculiarity at x equals 1. At x equals 1, the curve is continuous, but f prime of 1 does not exist. And this means that either the tangent is vertical or the secant lines disagree. Since we know the shape of the curve before and after x equals 1, this means we need to join the segments of the curve so that one of these two things happen. And possibly if we join them together like this, we can make that tangent line vertical. And we can join the other segments, keeping in mind that x equals 3 is a line we cannot cross. So we'll join the sections of our curve together to produce a graph that looks something like this. What if we have a different set of information? Since the limit as x goes to infinity of f of x is equal to 5, and we're graphing y equals f of x, we know that as x goes to infinity, our y values go to 5. And this gives us a horizontal asymptote of y equals 5, but it's useful to remember we only need this for the far, far, far right of the graph. Similarly, since as x goes to minus infinity f of x is equal to 3, then as x goes to minus infinity, y goes to 3. And this gives us a horizontal asymptote of y equals 3, but only for the far, far, far, far left of the graph. Now since f of x is continuous everywhere, there are no lines that we can't cross. However, since f prime of 5 is undefined, something odd happens at x equals 5. So let's set down a reminder line. We can cross this line. That's not a problem, but we have to remember that something strange is going to happen here. So let's piece things together. For x less than 5, the derivative is positive, and the second derivative is positive. And so we know the tangent line slopes upward, and the tangent line is below the curve. So we can draw a placeholder point and the curve through that point, and again we know we're going to have to adjust the location and shape of this curve slightly. For x greater than 5, f prime of x is negative, and f double prime is positive, and so we know the tangent line slopes downward, and the tangent line is below the curve, and so our curve will look something like this. And since f prime of 5 is undefined, we know the tangent line is either vertical or fails to exist, but since f of x is continuous through x equals 5, we do know that our curve joins across this line. So we just have to make sure that when we join our segments together, we either get a vertical tangent or we get a tangent line that doesn't exist because the secant lines disagree. And so maybe we'll join them something like this.