Now that we have a visual idea of what it means for a function to be differentiable at a point, let’s get a little more formal. We can call a function differentiable at a single point, over an interval, or over its entire domain.
• A function is differentiable at a point if and only if the function has a derivative, as defined at the end of Chap. 3, at that point.
• A function is differentiable over an interval if and only if it has a derivative, as defined at the end of Chap. 3, at every point in that interval.
• A function is differentiable in general (or simply differentiable) if and only if it has a derivative, as defined at the end of Chap. 3, at every point in its domain.
Try to find the limits
When we want to know whether or not a function is differentiable at a particular point, we can try to find the derivative at that point and see if the result makes sense. If we get a meaningful and unambiguous result, then the derivative exists at the point. Otherwise, it doesn’t.
For all the functions f (x ) we examined in Chap. 3, we were able to find the following limit at any point where the independent variable was equal to some specific value x0:
f ′(x0)= Lim
Δ →0x [ f (x0+ Δx ) −f (x0)] / Δx
If the notion of a derivative is to make any sense for a function f (x ) at the point where x=x0, the value x0 must be in the domain. That is to say, f (x0) must be defined. Once we know that, then we must be able to find the above limit whether Δx is positive or negative. In other words, the limit has to exist as we approach x0 from the right, and it also has to exist as we approach x0 from the left. That’s not all! For f (x ) to have a derivative at the point where x = x0, the right-hand and left-hand limits, as defined in Chap. 2, must be the same. Stated symbolically,
Figure 4-4 is a graph of a quadratic function that is defined and continuous over the entire set of real numbers:
f (x ) =x2− 3
When We Can Differentiate 59
Let’s see whether or not this function is differentiable at the point where x= −2.5. We set up the limit
Limx
Δ →0 [ f (−2.5+ Δx ) −f (−2.5)] / Δx We can calculate f (−2.5) easily enough:
f (−2.5)= (−2.5)2− 3
= 6.25 − 3
= 3.25 Now we can write the limit as
Limx
Δ →0 [(−2.5+ Δx )2− 3 − 3.25] / Δx which expands to
Limx
Δ →0 [6.25 − 5Δx+ (Δx )2− 3 − 3.25] / Δx and then simplifies to
Limx
Δ →0 [−5Δx+ (Δx )2] / Δx
x f(x)
2 4 6
–2 –4 –6
2 6
–6 4
f(x) = x2 – 3 –4
x= –2.5
f¢(–2.5) = –5
Figure 4-4 This function is differentiable at the point where x= −2.5.
and finally to
Limx
Δ →0 −5+ Δx
WhetherΔx is positive or negative (that is, whether Δx approaches 0 from the right or the left), it’s easy to see that this limit is equal to −5. Therefore,
f ′(−2.5)= −5
Now that we’ve found the derivative at the point where x= −2.5, we can conclude that the function is differentiable there.
Another example
Let’s reexamine the function we saw in Fig. 4-2. Suppose we want to find out if this function is differentiable at the point where x = 3. In the vicinity of this point, we can imagine that the function is
f (x ) = 3
as shown in Fig. 4-5. To see whether or not this function is differentiable at the point where x= 3, we can try to find the derivative there by setting up the limit
Limx
Δ →0 [ f (3 + Δx ) −f (3)] / Δx
When We Can Differentiate 61
x
Figure 4-5 This function is differentiable at the point where x = 3, but not where x= 0.
We know that f (3) = 3, so we can rewrite this limit as Limx
Δ →0 [ f (3 + Δx ) − 3] / Δx
We must keep in mind that this limit has to be evaluated as Δx becomes very small either posi-tively or negaposi-tively, approaching 0 from either the right or the left. In either of those scenarios, the value of f (3 + Δx ) remains constant at 3. That allows us to simplify the above limit to
This limit is equal to 0. No matter how small we make Δx, it’s still a tiny positive or negative real number (never 0), so 0/Δx is always equal to 0. Therefore
f ′(3)= 0
Having found a well-defined derivative at the point where x= 3, we can be sure that the func-tion is differentiable there.
Are you confused?
“Wait!” you say. “In the two examples we just finished, we didn’t find the left-hand and the right-hand limits separately! Doesn’t the definition of the derivative, given at the end of Chap. 3, require us to do that?” Yes, it does. But in the two examples we just finished, we were able to perform both tasks together, because it was apparent that the two limits would be the same if we calculated them individually. If your mind demands absolute rigor (not a bad trait for a mathematician), you can do both limit calculations separately in every case. In the following challenge, we’ll see a situation where that attitude serves us well.
Here’s a challenge!
Let’s try to find the derivative of the function shown in Fig. 4-5 at the point where x= 0, first by approaching the point from the right, and then by approaching from the left. What can we conclude from this result?
Solution
Before we start, let’s state the function mathematically. To fully describe it, we must write it in two parts, like this:
f(x ) = −3 when x< 0
= 3 when x≥ 0
First, let’s approach 0 from the right, that is, from the positive side. Geometrically, we’ll move from right to left along the upper half-line, or the portion to the right of the vertical axis. We set up the limit
Limx
Δ → +0 [ f (0 + Δx ) −f (0)] / Δx
We know that f (0) = 3, so we can rewrite this limit as
This result makes sense. The right-hand part of the half-line is horizontal, so we should expect the deriva-tive, which represents the slope, to be 0.
Now let’s approach 0 from the left (from the negative side). In the graph, this is the equivalent of moving from left to right along the lower half-line, which lies to the left of the vertical axis. We set up the limit
Limx
AsΔx becomes smaller and smaller negatively, f (0 + Δx ) remains constant, maintaining a value of −3 (not 3, as it does to the right of the vertical axis), so we can rewrite the above limit as
Limx
Here, the value of Δx becomes smaller and smaller negatively, becoming arbitrarily close to 0, but never quite getting there. The ratio −6/Δx is always a positive real, because it’s a negative divided by a negative.
AsΔx approaches 0, the ratio −6/Δx blows up; it gets arbitrarily large. The limit is not defined. This result tells us that the function
f(x ) = −3 when x< 0
= 3 when x≥ 0 is nondifferentiable at the point where x= 0.