Intermediate Value Property and Limits of Derivatives

The Intermediate Value Theorem says that if a function is continuous on an interval, then it has the intermediate value property on that interval. That is, if f is continuous on the interval I, and a; b 2 I, then for any K between f .a/ and f .b/, there is a c between a and b with f.c/ D K. Suppose that f is differentiable at each point of an interval I. If f⁰ is continuous on I, then certainly it obeys the Intermediate Value Theorem and has the intermediate value property on I. But f⁰.x/ can exist for all x 2 I without f⁰ being a continuous function. One example is f.x/ D

x²sin₁

x2

 if x ¤0 0 if x D0



. This function is differentiable for all x. When x ¤ 0, the derivative is f⁰.x/ D 2x sin 0, f⁰is not even bounded and, in fact, oscillates wildly. In spite of its discontinuity at 0, f⁰does have the intermediate value property. For example, for any x ¤0, the function f⁰obtains every value between f⁰.x/ and f⁰.0/ D 0 on the interval between

156 5 Derivatives

Fig. 5.7 x²sin₁

x²

and its derivative

0 and x. Moreover, it obtains each of those values infinitely often. In fact, between 0 and x, the function f⁰takes on every real number infinitely often (Fig.5.7).

Note that the function f.x/ C x has a derivative of 1 at x D 0. This is an example of a function with a positive derivative at 0 which is not an increasing function over any open interval containing 0. This can easily be seen by the fact that in every open interval containing 0 there are intervals where the derivative of f.x/ C x is negative.

So, how can you prove that if a function f has a derivative f⁰ on an interval I, that f⁰ has the intermediate value property on I? The hypothesis suggests that you start by taking a function f differentiable on an interval I and values a; b 2 I. Then you select a value K between f⁰.a/ and f⁰.b/. Without loss of generality, you can assume that a < b and f⁰.a/ < K < f⁰.b/. The goal would be to show that there is a c between a and b such that f⁰.c/ D K. One simplification is to replace f with the function g.x/ D f .x/  Kx. This function is also differentiable on I, and if f⁰.c/ D K, then g⁰.c/ D 0. Which theorems about derivatives allow you to conclude that a derivative is 0 at some point in an interval? First there is a theorem that states that if a differentiable function reaches an extreme value at a point in an interval, then the point is either a critical point of the function or an endpoint of the interval.

A second theorem is Rolle’s Theorem which talks about a differentiable function which takes on the same value at the endpoints a and b. Since you do not have any information about the values of g at the endpoints of the interval, the theorem about extreme values may be the more promising choice for this proof.

What is known about the function g? You know that g is differentiable at each point of the interval from a to b. Additionally, g⁰.a/ D f⁰.a/  K < 0 and g⁰.b/ D f⁰.b/  K > 0. Does this mean that the function g is decreasing at a and increasing at b? Well, it would if you knew that g⁰ were continuous because then g⁰would be negative in an interval around a and positive in an interval around b. But, as you now know, g⁰need not be continuous. On the other hand, there is a theorem that says that if g⁰.a/ is negative, then there is a ı > 0 such that if x satisfies a < x < a C ı, then g.x/ < g.a/. This does not show much, but you can use it to conclude that g does not take on its minimum value onŒa; b at a. A similar argument uses the fact that g⁰.b/ > 0 to show that g does not take on its minimum value on Œa; b at b.

Is g a continuous function onŒa; b? It is differentiable at each point of Œa; b, so it is continuous. All continuous functions on a close bounded interval take on both their minimum and maximum values on the interval. Thus, you know that g takes on its minimum value onŒa; b at some point c strictly between a and b. Such a point must be a critical point of g, so g⁰.c/ D 0. This is the idea behind the following proof.

PROOF: A function differentiable on an interval has the intermediate value property on that interval.

• Let f be a function differentiable at each point of an interval I.

• Let a; b 2 I, and assume that f⁰.a/ ¤ f⁰.b/.

• Without loss of generality assume that a< b and f⁰.a/ < f⁰.b/.

• Let K be a value satisfying f⁰.a/ < K < f⁰.b/.

• Let g.x/ D f .x/  Kx.

• Then g⁰.x/ D f⁰.x/  K for all x 2 Œa; b, and g⁰.a/ < 0 < g⁰.b/.

• Since g is differentiable at each point ofŒa; b, it is continuous on Œa; b.

• Since g is continuous onŒa; b, it obtains a minimum value at some point c2 Œa; b.

• g⁰.a/ < 0 implies that there is a ıa > 0 such that g.x/ < g.a/ for all x satisfying a< x < a C ıa. In particular, g does not obtain its minimum at a.

• g⁰.b/ > 0 implies that there is a ıb > 0 such that g.x/ < g.b/ for all x satisfying b ıb< x < b. In particular, g does not obtain its minimum at b.

• It follows that g obtains its minimum onŒa; b at a point c strictly between a and b.

• Since c is not an endpoint ofŒa; b, g⁰.c/ D 0.

• Thus, f⁰.c/ D g⁰.c/CK D K which shows that f⁰has the intermediate value property on I.

There are simple examples of functions that have discontinuous derivatives that do not have the intermediate value property; functions such as f.x/ D jxj. This function’s derivative is the constant 1 for all x > 0 and 1 for all x < 0. This derivative is not continuous at x D0 because it is not defined there. Clearly, f⁰does not have the intermediate value property on any interval containing both positive and negative numbers, but then f does not satisfy the hypothesis of the previous theorem on any such interval because f⁰.0/ is not defined. Functions that have discontinuous derivatives that are defined at all points will have to exhibit wild oscillations in the neighborhoods of those discontinuities similar to the example

x²sin₁

x²

 if x ¤0 0 if x D0

 .

Suppose f is a function whose derivative is defined at all points of an interval except perhaps at some point c in the interval. What can be said if lim

x!cf⁰.x/ exists?

Such a derivative does not exhibit wild oscillations near c, and, in fact, it must have a continuous derivative at c. The proof is a consequence of the Mean Value Theorem.

158 5 Derivatives

PROOF: Let f be a function differentiable at all points of an interval.a; b/

except perhaps at some point c2 .a; b/. Suppose that the limit lim_x!cf⁰.x/

exists. Then f⁰is continuous at c.

• Let f be a function differentiable on the interval .a; b/ except perhaps at c2 .a; b/.

• Assume lim

x!cf⁰.x/ D L.

• From the definition of limit, given  > 0, there is a ı > 0 such that if y2 .a; b/ with 0 < jy  cj < ı, then jf⁰.y/  Lj < .

• Let x 2.a; b/ with 0 < jx  cj < ı.

• By the Mean Value Theorem there is a y between x and c such that

f.x/f .c/

xc D f⁰.y/.

• Then y 2.a; b/ with 0 < jy  cj < ı, soˇˇˇ^{f.x/f .c/}_xc  Lˇˇˇ D jf⁰.y/  Lj < .

• Thus, lim

x!c f.x/f .c/

xc D L, so f⁰.c/ D L.

• Because f⁰.c/ D lim_x

!cf⁰.x/, it follows that f⁰.x/ is continuous at c.

• This completes the proof.