.
Also we have for all x in [a, b], f (x)
-
g x
³ f (b)
-
g b, that is, f (b)
- f (x)
£
g b
- g
x. Consequently, for all x in [a, b), ( f (b)
- f (x))/(b
- x) £
g since b
- x >
for x < b. Similarly since f is differentiable at b,
Intermediate Value Theorem for the Derived Function (Darboux's Theorem)
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The intermediate value theorem for continuous function is a very useful result. A lesser known result about differentiable function is that its derived function also has the intermediate value property.
Theorem 1. Let I be an open interval and suppose f : I ® R is a differentiable function. Let a, b be two points in I such that a < b. Suppose f ' (a) ¹ f ' (b). Then for any value g strictly between f ' (a) and f ' (b), there is a point c in (a, b) such that f ' (c) = g.
Proof. Let us define the following function g: I ® R by g(x) = f (x) - g x for x in I. Then g is differentiable and g' (x) = f ' (x) - g . If we can show that g has either a relative maximum or a relative minimum at a point c in (a, b), then we are done. Consider the function h : [a, b] ® R, the restriction of g to the closed interval, [a, b]. Then since g is differentiable, g is also continuous on [a, b]. Therefore, by the Extreme Value Theorem, the restriction of g, h attains both its maximum and minimum in [a, b]. We shall show that at least one of the absolute maximum or absolute minimum occurs in the interior of [a, b]. Suppose h(a) is the maximum and h(b) is the minimum. Then for all x in [a, b], h(x) £ h(a) and h(x) ³ h(b). Hence for all x in [a, b], f (x) - g x £ f (a) - g a, that is, f (x) - f (a) £ g x - g a. Therefore, for all x in [a, b] and x ¹ a, ( f (x) - f (a))/(x - a) £ g . Since f is differentiable at a,
.
Also we have for all x in [a, b], f (x)
-
g x
³ f (b)
-
g b, that is, f (b)
- f (x)
£
g b
- g
x. Consequently, for all x in [a, b), ( f (b)
- f (x))/(b
- x) £
g since b
- x >
for x < b. Similarly since f is differentiable at b,
.
Therefore, we can conclude that f ' (a) and
f ' (b) are both less than or equal to
g, contradicting that
g is strictly between
f ' (a) and f ' (b).
Similarly, if h(a) is the minimum and
h(b) is the maximum, we can show in like
manner that f ' (a) and f '
(b) are both greater than or equal to g,
giving a contradiction to that g
is strictly between f ' (a) and
f ' (b). We have thus shown that one of the maximum or minimum of
h must occur at a point c in (a, b). Since h(c)
is also a relative extremum and h is differentiable, h ' (c)
= f ' (c) -
g = 0 and so f
' (c) =
g . (See Theorem
6.1.2, Calculus, an introduction page 82.) This completes the proof.
Corollary 2. Let I be an open interval and suppose f : I ® R is a differentiable function. Then the image of the derived function of f , f ' (I ) is also an interval.
Proof. The proof is similar to Theorem 4.6.12 of Calculus, an introduction, page 53 that is the continuous image of an interval is an interval. Let f ' (a) ¹ f ' (b) be in f ' (I ). We may assume that f ' (a) < f ' (b). Theorem 1 says that for any g such that f ' (a) < g < f ' (b) , g Î f ' (I ). Hence the interval [ f ' (a), f ' (b)] Í f ' (I ). Therefore, by the usual characterisation of an interval, f ' (I ) is an interval.
Remark 3. In the proof of Theorem 1 above, only the right derivative of f at x = a and the left derivative of f at b is used. Therefore, if the domain of f is an interval I not necessarily open and if derivative of f is appropriately defined as the right derivative or left derivative at the end point or end points of its domain I, then Theorem 1 holds true when I is just any interval. Consequently, Corollary 2 is also true when the domain of f , I , is replaced by any interval, not necessarily open.
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