The “challenge” we just finished involves a function that can be differentiated everywhere except at a single point. The problem in this particular case is caused by a discontinuity at the point where x= 0. There are other ways that a function can be nondifferentiable at a point.
When We Can’t Differentiate 63
Example
Let’s consider another function with a discontinuity, but of a different sort than the one in the
“challenge” we solved a moment ago. Figure 4-6 is a graph of the modified reciprocal function f(x ) = 1/x when x≠ 0
= 0 when x= 0
This function is defined over the entire set of real numbers, but it has a discontinuity at (0,0).
If we want to find out whether or not f (x ) is differentiable at the point where x = 0, we must set up the limit
Limx
Δ →0 [ f (0 + Δx ) −f (0)] / Δx
and evaluate it both from the right and from the left. Let’s do it from the right first. We write this as
Limx
Δ → +0 [ f (0 + Δx ) −f (0)] / Δx which simplifies to
Limx
Δ → +0 [ f (Δx ) −f (0)] / Δx
x f(x)
f(x) = 1/xwhenx≠ 0 = 0 when x= 0
2 4 6
–4 –6
2 6
–4
–6 4 (0,0)
Figure 4-6 This function isn’t continuous at the point where x= 0, and it isn’t differentiable there, either.
because 0 + Δx= Δx. We are told that f (0) = 0, so we can simplify further, getting Limx
Δ → +0 f (Δx ) / Δx
The function tells us to take the reciprocal of whatever nonzero number we put in. That means we can rewrite the above expression as
Limx
Δ → +0 [1/(Δx )] / Δx which simplifies to
Limx
Δ → +0 1/(Δx )2
AsΔx approaches 0 from the right (the positive side), the ratio 1/(Δx )2 grows arbitrarily large in the positive sense. That’s because we’re taking the reciprocal of (Δx )2, an endlessly shrinking positive real. The above limit blows up positively; it’s not defined.
The same thing happens when we approach 0 from the left. We can rework the sequence of steps, replacing Δx→ 0+ with Δx→ 0− in every instance, to arrive at
Limx
Δ → −0 1/(Δx )2
This limit blows up positively, just as the right-hand limit does, so it’s undefined as well. Actu-ally, this second exercise is overkill. Once we showed that the right-hand limit doesn’t exist at the point where x= 0, we gathered enough information to know that our function isn’t differentiable there.
It’s tempting to say that the discontinuity here is somehow “worse” than the one in the func-tion shown by Fig. 4-5. In the earlier case, a limit could be defined from one side, at least. This time, there is no definable limit from either side. We can be “double-sure” that the function
f(x ) = 1/x when x≠ 0
= 0 when x= 0 is nondifferentiable at the point where x= 0.
Another example
Figure 4-7 is a graph of the absolute-value function, which can be written in two-part fashion like this:
f(x ) =x when x≥ 0 = −x when x< 0
This function is defined and continuous over the entire set of reals. If x> 0 or x< 0, the absolute-value function behaves as a linear function, and it is therefore differentiable when we restrict the domain to either the set of strictly positive reals or the set of strictly negative reals.
As we can see by examining Fig. 4-7, the graph turns a corner. The slope changes sud-denly from −1 to 1 as we move to the right through the point where x= 0. When the slope of
When We Can’t Differentiate 65
a graph changes abruptly at a point, we should suspect that point as a likely place where the function is nondifferentiable! Let’s find out if that’s the case here. First, we’ll evaluate the limit of the slope as we approach the point where x= 0 from the right:
Limx
Δ → +0 [ f (0 + Δx ) −f (0)] / Δx Because 0 + Δx= Δx, this can be simplified to
Limx
Δ → +0 [ f (Δx ) −f (0)] / Δx We can simplify further, knowing that f (0) = 0:
Limx Obviously, Δx/Δx= 1. That means we have
Limx
Figure 4-7 This function is continuous at the point where x= 0, but it isn’t differentiable there.
This limit is equal to 1. We can believe this easily enough. The right-hand half of the graph is a half-line with a slope of 1.
Now, let’s take the limit of the slope as we approach the point where x= 0 from the left.
We write Knowing that f (0) = 0, we can substitute and rewrite this as
Limx
Δ → −0 f (Δx ) / Δx
When we take the absolute value of a negative number, the function simply reverses the sign of the number, so when Δx is small and negative, f (Δx ) = −Δx. That means we can simplify the limit some more, writing it as
Limx
Δ → −0 −Δx/Δx The ratio −Δx/Δx is always equal to −1, so we have
Limx Δ → −0 −1
This is equal to −1, a sensible result. The left-hand part of the graph is a half-line with a slope of−1.
We have found the limits in the prescribed form, both from the right and from the left.
But they aren’t the same, showing that the function f (x ) = |x|
is nondifferentiable at the point where x= 0. (The vertical lines in this equation indicate that the absolute value is to be taken of whatever quantity appears between them.)
Are you confused?
“I see what’s going on now,” you say. “If the slope of a graph changes suddenly at a point, then even if the function is continuous, there’s no derivative at that point. If we approach the point from one side, the slope approaches a certain value, but if we approach the point from the other side, the slope approaches a different value. A derivative can’t be defined as more than one slope at a time. Is that right?” Yes, that geo-metrically describes the situation. It’s not rigorous, but it’s a good way to talk about it informally.
Then you ask, “What about a function where two different curves terminate and meet at a single point, but their slopes approach the same value at the point? Is such a two-part function differentiable at the point where the curves come together?” That is a great question! You’ll learn the answer (in one case, anyway ) as you work out Exercise 10 at the end of this chapter.
When We Can’t Differentiate 67
Here’s a challenge!
Look once again at the function graphed in Fig. 4-3. Show that it’s nondifferentiable at the point where x= 2.
Solution
There are two components to the function. When the input is smaller than 2, the output is the square of the input. When the input is 2 or larger, the output has a constant value of 4. We can express this situa-tion by writing
f(x ) =x 2 when x< 2
= 4 when x≥ 2
The function is defined and continuous over the set of real numbers, but the slope of its graph abruptly changes at the point where x= 2. We can attempt to find the derivative at the point where x= 2 by evaluating
Limx
Δ →0 [ f (2 + Δx ) −f (2)] / Δx from both the right and the left. Let’s do it from the right first. We have
Limx
Δ → +0 [ f (2 + Δx ) −f (2)] / Δx
WhenΔx is small and positive, we’re in the part of the function with a constant value of 4. We’re told thatf (2) = 4. When we substitute 4 in place of f (2 + Δx ), and substitute 4 in place of f (2) in the above
This limit is obviously equal to 0. When we examine the graph, that conclusion seems reasonable, because the part of the graph to the right of the point where x= 2 appears as a straight, horizontal line.
Now let’s go to the left of the point where x= 2, into the part of the graph that has a parabolic shape.
WhenΔx is small and negative, we’re in the zone where the function squares the input. We now must work with the limit
Limx
Δ → −0 [f (2 + Δx ) −f (2)] / Δx
We must not forget that f (2) = 4. When we substitute (2 + Δx )2 for f (2 + Δx ), and substitute 4 for f (2) in the above expression, it becomes
Limx
Δ → −0 [(2+ Δx )2− 4] / Δx
When we square the binomial, we get
This limit is equal to 4. That’s a sensible outcome. The slope of the parabola, if it were complete, would be 4 at the point where x= 2. But that’s not the same value as we got when we evaluated the right-hand limit.
Having found a disagreement between the right-hand and left-hand limits of the slope, we know that this function is nondifferentiable at the point where x= 2.
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems.
Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it!
1. Sketch a graph of the following function:
f(x ) = 1/x when x< −1 = −1 when x≥ −1
At what point or points does it appear, based on the graph, that this function is nondifferentiable?
2. Verify the answer to Prob. 1 mathematically.
3. Sketch a graph of the following function:
f (x ) =x when x≤ 3 = 1 when x> 3
At what point or points does it appear, based on the graph, that this function is nondifferentiable?
4. Verify the answer to Prob. 3 mathematically.
5. Sketch a graph of the following function, noting the subtle difference between it and the function described in Prob. 3:
f (x ) =x when x< 3 = 1 when x≥ 3
Practice Exercises 69
At what point or points does it appear, based on the graph, that this function is nondifferentiable?
6. Verify the answer to Prob. 5 mathematically.
7. Sketch a graph of the following function:
f (x ) =x 2 when x< 1 =x 3 when x≥ 1
At what point or points does it appear, based on the graph, that this function is nondifferentiable?
8. Verify the answer to Prob. 7 mathematically.
9. Sketch a graph of the following function:
f (x ) =x 2 when x< 1 = 2x− 1 when x≥ 1
At what point or points does it appear, based on the graph, that this function is nondifferentiable?
10. Verify the answer to Prob. 9 mathematically.
In this chapter, we’ll learn how to differentiate a function that raises a variable to a large inte-ger power. We’ll learn how to differentiate the sum of two functions. Then we’ll combine these rules to differentiate functions that can be written in polynomial form.