56
x
f(x) = 1/x
2 4 6
–4 –6
2 6
–4
–6 4
“Gap” at x= 0
Figure 4-1 This function has a gap at x= 0, because it is not defined there. It is discontinuous at the point where x= 0.
x f(x)
2 4 6
–4 –6
2 6
–4
–6 4
f(x) = –3 when x< 0 = 3 when x≥ 0
“Jump” at x= 0
–2
Figure 4-2 This function takes a jump at x= 0, even though it is defined there. It is discontinuous at the point where x= 0.
We might think that the slope at the point (0,3) is parallel to the x axis, running along the part of the graph that lies to the right of that point. After all, the whole graph looks horizontal.
I might try to convince you, based on that notion, that the function has a slope of 0 at every point in its domain, including (0,3). But you could argue that a tangent line at the jump is not horizontal but vertical, running along the f (x ) axis and passing through (0,−3) as well as (0,3). That makes just as much (or just as little) sense.
The slope of this graph at the point where x= 0 can’t be defined with certainty. Therefore, the derivative does not exist for this function at the point where x= 0. Both this function and the one with the gap (Fig. 4-1) have discontinuities. As things work out, if a function is discontinuous at a particular point, then it has no derivative there. Turning this logic “inside-out” gives us one of the most important facts in calculus:
• If a function has a derivative at a point, then the function is continuous there.
Is there a corner?
Figure 4-3 shows the graph of a function that’s defined over the entire set of real numbers, and that’s continuous everywhere as well. But there’s something strange about this function at the point where x= 2. It has no gap there, and its value doesn’t jump there, but the curve turns a corner.
The slope of a tangent line can’t be defined at the point where x = 2. The trouble here is similar to the problem we encountered in Fig. 4-2. If we want to talk about the slope at the point (2,4) on this graph, should we base our idea on the half-parabola to the left, or on the half-line to the right? Neither idea is better, but they disagree.
Let’s Look at the Graph 57 although it is continuous over the entire set of real numbers.
Are you confused?
You might still wonder why we can’t define the slope of the graph in Fig. 4-3 at the point (2,4). You might say, “Suppose we choose two movable points, one on either side of (2,4), and draw a line through them.
As we move these two points closer and closer to (2,4), one along the half-parabola and the other along the half-line, won’t they approach a line having a slope of 2 and running through the point (2,4)? Won’t that line be a legitimate tangent line?” No, there are two problems with that idea.
First, it’s “illegal” to use two movable points in an attempt to find a line whose slope indicates a derivative.
That’s not the way it’s done according to the definition at the end of Chap. 3. We must approach the fixed point, in this case (2,4), from one side at a time, and use it as one of the points through which we draw the line. Then we must be sure that we end up with the same line when we approach from the left as we do when we approach from the right. Only then can we assign it a slope and determine the derivative based on that slope.
Second, even if it were “legal” to use two movable points to find the line, the slope of that line would depend on the “relative rates of approach” in a situation like the one we see in Fig. 4-3. If the point on the left is always twice as far away from (2,4) as the point on the right, for example, we will end up with a different line than we will get if the two points are always at the same distance from (2,4). By adjusting the
“relative rates of approach” of the two points, we could get a line with any slope between 0 and 4!
Here’s a challenge!
Look again at the graph of the function f (x ) = 1/x, shown in Fig. 4-1. Can we tell, by visual inspection, what happens to f ′(x ) as x increases endlessly (that is, “approaches positive infinity”)? Can we tell what happens to f ′(x ) as x decreases endlessly (that is, “approaches negative infinity”)?
Solution
Yes, we can get an idea of what happens in these cases, although looking at a graph doesn’t constitute a math-ematical proof of anything. In this particular graph, as x increases endlessly (that is, “approaches positive infin-ity”), the slope of the curve, and therefore the derivative f ′(x ), appears to approach 0 from the negative direction.
Asx decreases endlessly (“approaches negative infinity”), the slope of the curve, and therefore the derivative, again appears to approach 0 from the negative direction. We can write these statements symbolically as
Asx→ +∞,f ′(x ) → 0− and
Asx→ −∞,f ′(x ) → 0−
Don’t get confused here. The function itself behaves quite differently than its derivative! As x increases endlessly in the positive direction, the value of the function approaches 0 from the positive direction. As x increases endlessly in the negative direction, the value of the function approaches 0 from the negative direc-tion. We can write these statements symbolically as
Asx→ +∞,f (x ) → 0+
and
Asx→ −∞,f (x ) → 0−