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Next: New Functions from old Up: Functions, continuity and differentiability Previous: Definitions

Properties

We show the important properties called the intermediate value property and extremal value property of continuous functions. We also deduce the clutch of theorems called mean value theorem, Rolle's theorem and so on for differentiable functions.

The following important property of continuous functions will be used all the time.

Exercise 39   Let f (x) be a continuous function and {xn} be a sequence converging to c, then {f (xn} is a sequence converging to f (c). (Hint: Examine the condition for continuity near c).

Let f (x) be continuous for x satisfying a $ \leq$ x $ \leq$ b. Let c be a real number lying between f (a) and f (b) we want to show that c is a value of f; in other words any number intermediate to two values is itself a value.

Exercise 40   Let s be the least upper bound of the set

{x| a $\displaystyle \leq$ x $\displaystyle \leq$ b andf (x) $\displaystyle \leq$ c}

Show that f (s) = c.(Hint: To show that f (s) $ \geq$ c take a sequence of points approaching s from above).

Now let C be the least upper bound of the values of f (x), i. e. it is the least upper bound of the set {f (x)| a $ \leq$ x $ \leq$ b}. The C is an extremal value for f.

Exercise 41   Show that C = f (x) for some x in the range a $ \leq$ x $ \leq$ b. (Hint: We have a sequence {xn} so that f (xn) converges to C; by the section on sequences this sequence has a convergent subsequence).

The following property of differentiable functions is very important

Exercise 42   Let f (x) be differentiable for x satisfying a $ \leq$ x $ \leq$ b. Let s be such that f (s) is an extremal value for f. Then f'(s) = 0. (Hint: Examine the condition for differentiability near s).

Now suppose that f (x) is differentiable in the range a < x < b and continuous at the endpoints a and b as well. Suppose that f (a) = f (b) = 0. There is a point s where f attains its maximal value; similarly there is a point t where f attains its minimal value. If s is the point a or b then f (x) $ \leq$ 0 and if t is the point a or b then f (x) $ \geq$ 0. Thus in case both of these occur then f (x) = 0 for all x; then let c = (a + b)/2. Otherwise let c be any one of s and t which is not a or b. Thus we have a point where f'(c) = 0.

Exercise 43   For a general function g(x) which is differentiable in the range a < x < b and continuous at the endpoints we apply this to the function

f (x) = g(x) + $\displaystyle {\frac{g(b) - g(a)}{b-a}}$ . (x - a)

to show that there is a point c where

g'(c) = $\displaystyle {\frac{g(b) - g(a)}{b-a}}$


next up previous
Next: New Functions from old Up: Functions, continuity and differentiability Previous: Definitions
Kapil H. Paranjape 2001-01-20