Continuity. MATH 161 Calculus I. J. Robert Buchanan. Fall Department of Mathematics

Transcription

1 Continuity MATH 161 Calculus I J. Robert Buchanan Department of Mathematics Fall 2017

2 Intuitive Idea A process or an item can be described as continuous if it exists without interruption. The mathematical definition of continuity is more technical than intuitive, but is based on the concept of the limit.

3 Graphical Examples of Discontinuities (1 of 4) y x a f (a) is undefined. The graph has a hole at x = a.

4 Graphical Examples of Discontinuities (2 of 4) y a x f (a) is defined but the lim x a f (x) does not exist. This is sometimes called a jump discontinuity.

5 Graphical Examples of Discontinuities (3 of 4) y f(a) f (a) is defined and the lim x a f (x) exists, but lim f (x) f (a). x a x a

6 Graphical Examples of Discontinuities (4 of 4) y a x The lim x a f (x) does not exist (is not finite).

7 Definition of Continuity Definition A function f is continuous at x = a when 1. f (a) is defined, 2. lim x a f (x) exists, and 3. lim x a f (x) = f (a). Otherwise f is said to be discontinuous at x = a.

8 Examples (1 of 2) Determine the values of x for which f (x) = x 2 1 x 1 is continuous.

9 Examples (2 of 2) Consider the piecewise-defined function { ce x + 1 if x < 0, f (x) = sin 1 ( x 2) if 0 x 2. Determine a value for the constant c which makes f (x) continuous at x = 0.

10 Removable Discontinuities If we can eliminate a discontinuity at x = a by redefining the function at x = a, then we say the discontinuity is removable.

11 Removable Discontinuities If we can eliminate a discontinuity at x = a by redefining the function at x = a, then we say the discontinuity is removable. Example If f (x) = sin x for x 0, is the discontinuity at x = 0 x removable? If so, define f (0) so that f (x) is continuous at x = 0.

12 Solution f(x)= sin x x x

13 Solution f(x)= sin x x x sin x if x 0, g(x) = x 1 if x = 0.

14 Some Continuous Functions Theorem All polynomials are continuous for all x. Likewise sin x, cos x, tan 1 x, and e x are continuous for all x. The radical function n x is continuous for all x when n is odd and continuous for all x > 0 when n is even. Also ln x is continuous for all x > 0 and sin 1 x and cos 1 x are continuous for 1 < x < 1.

15 Arithmetic of Continuous Functions Theorem Suppose f and g are continuous at x = a, then 1. (f ± g) is continuous at x = a, 2. (f g) is continuous at x = a, ( ) f 3. is continuous at x = a, provided g(a) 0. g

16 Example Determine the values of x where f (x) = discontinuous. 4x x 2 + x 2 is

17 Example Determine the values of x where f (x) = discontinuous. 4x x 2 + x 2 is Strategy: to find discontinuities, look for values of x for which f (x) is undefined, approaches an infinite limit, or has differing one-sided limits.

18 Example Determine the values of x where f (x) = discontinuous. 4x x 2 + x 2 is Strategy: to find discontinuities, look for values of x for which f (x) is undefined, approaches an infinite limit, or has differing one-sided limits. Function f (x) is continuous for x (, 2) ( 2, 1) (1, ).

19 Composition of Functions and Continuity Theorem Suppose that lim g(x) = L and that f is continuous at x = L, x a then ( ) lim f (g(x)) = f lim g(x) = f (L). x a x a

20 Composition of Functions and Continuity Theorem Suppose that lim g(x) = L and that f is continuous at x = L, x a then ( ) lim f (g(x)) = f lim g(x) = f (L). x a x a Corollary Suppose that g is continuous at x = a and that f is continuous at x = g(a), then f g is continuous at x = a.

21 Example Find the values of x for which h(x) = ln(4 x 2 ) is continuous.

22 Example Find the values of x for which h(x) = ln(4 x 2 ) is continuous. Let h(x) = f (g(x)) where g(x) = 4 x 2 and f (x) = ln x. Function g(x) is continuous for all x and f (x) is continuous for all x > 0, thus h(x) is continuous at all x for which g(x) > 0.

23 Example Find the values of x for which h(x) = ln(4 x 2 ) is continuous. Let h(x) = f (g(x)) where g(x) = 4 x 2 and f (x) = ln x. Function g(x) is continuous for all x and f (x) is continuous for all x > 0, thus h(x) is continuous at all x for which g(x) > 0. Function h(x) is continuous for x ( 2, 2).

24 Continuity on an Interval Definition If f is continuous at every point on an open interval of the form (a, b) we say f is continuous on (a, b). We say f is continuous on [a, b] if f is continuous on (a, b) and lim x a f (x) = f (a) and lim f (x) = f (b). + x b If f is continuous on (, ) then we say that f is continuous everywhere.

25 Example Suppose the formula for the amount owed for state income tax is 0 if x 0, T (x) = a x if 0 < x < 20, 000, b (x 20, 000) if x 20, 000. where x is the income of the taxpayer. Determine values for the constants a and b that make the function T continuous everywhere.

26 Solution Since each piece of the piecewise-defined function is a polynomial, each piece is continuous. The transition points (x = 0 and x = 20, 000) are the only possible locations of discontinuities.

27 Solution Since each piece of the piecewise-defined function is a polynomial, each piece is continuous. The transition points (x = 0 and x = 20, 000) are the only possible locations of discontinuities. For x = 0, T (0) = 0 lim T (x) x 0 = lim x 0 T (x) + = lim lim x 0 x 0 thus T (x) is continuous at x = 0 if a = 0. +(a x) = a

28 Solution Since each piece of the piecewise-defined function is a polynomial, each piece is continuous. The transition points (x = 0 and x = 20, 000) are the only possible locations of discontinuities. For x = 0, T (0) = 0 lim T (x) x 0 = lim x 0 T (x) + = lim lim x 0 x 0 thus T (x) is continuous at x = 0 if a = 0. For x = 20, 000, T (20, 000) = b +(a x) = a lim T (x) = lim x 20,000 x 20,000 (a x) = 2, 400 lim x 20,000 T (x) = lim + +(b (x 20, 000)) = b x 0 thus T (x) is continuous at x = 20, 000 if b = 2, 400.

29 Intermediate Value Theorem (1 of 2) Theorem (Intermediate Value Theorem) Suppose f is continuous on the closed interval [a, b] and W is any number between f (a) and f (b). Then there is a number c [a, b] for which f (c) = W. f(b) W a f(a) c b

30 Intermediate Value Theorem (2 of 2) Corollary Suppose f is continuous on the closed interval [a, b] and f (a) and f (b) have opposite signs (in other words f (a)f (b) < 0). Then there is at least one number c (a, b) for which f (c) = 0.

31 Intermediate Value Theorem (2 of 2) Corollary Suppose f is continuous on the closed interval [a, b] and f (a) and f (b) have opposite signs (in other words f (a)f (b) < 0). Then there is at least one number c (a, b) for which f (c) = 0. This corollary forms the basis of the Method of Bisections for approximating the roots of equations.

32 Example (1 of 2) Use the Method of Bisections to approximate the solution to the equation 2 ln x x = 0. y x

33 Example (2 of 2) a b m f (a) f (b) f (m)

34 Homework Read Section 1.4 Exercises: 1 43 odd