Continuity

1. Continuity Alex Karassev

2. Definition • A function f is continuous at a number a if • Thus, we can use direct substitution to compute the limit of function that is continuous at a

3. Some remarks • Definition of continuity requires three things: • f(a) is defined (i.e. a is in the domain of f) • exists • Limit is equal to the value of the function • The graph of a continuous functions does not have any "gaps" or "jumps"

4. Continuous functions and limits • TheoremSuppose that f is continuous at band Then • Example

5. Properties of continuous functions • Suppose f and g are both continuous at a • Then f + g, f – g, fg are continuous at a • If, in addition, g(a) ≠ 0 then f/g is also continuous at a • Suppose that g is continuous at a and f is continuous at g(a). Then f(g(x)) is continuous at a.

6. Which functions are continuous? • Theorem • Polynomials, rational functions, root functions, power functions, trigonometric functions, exponential functions, logarithmic functions are continuous on their domains • All functions that can be obtained from the functions listed above using addition, subtraction, multiplication, division, and composition, are also continuous on their domains

7. Example • Determine, where is the following function continuous:

8. Solution • According to the previous theorem, we need to find domain of f • Conditions on x: x – 1 ≥ 0 and 2 – x >0 • Therefore x ≥ 1 and 2 > x • So 1 ≤ x < 2 • Thus f is continuous on [1,2)

9. Intermediate Value Theorem

10. River and Road

11. River and Road

12. Definitions • A solution of equation is also calledarootof equation • A number c such that f(c)=0 is calledarootof function f

13. Intermediate Value Theorem (IVT) • f is continuous on [a,b] • N is a number between f(a) and f(b) • i.e f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a) • then there exists at least one c in [a,b] s.t. f(c) = N y y = f(x) f(b) N f(a) x c a b

14. Intermediate Value Theorem (IVT) • f is continuous on [a,b] • N is a number between f(a) and f(b) • i.e f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a) • then there exists at least one c in [a,b] s.t. f(c) = N y y = f(x) f(b) N f(a) x c3 c1 c2 a b

15. Equivalent statement of IVT • f is continuous on [a,b] • N is a number between f(a) and f(b), i.e f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a) • then f(a) – N ≤ N – N ≤ f(b) – N or f(b) – N ≤ N – N ≤ f(a) – N • so f(a) – N ≤ 0 ≤ f(b) – N or f(b) – N ≤ 0 ≤ f(a) – N • Instead of f(x) we can consider g(x) = f(x) – N • so g(a) ≤ 0 ≤ g(b) or g(b) ≤ 0 ≤ g(a) • There exists at least one c in [a,b] such that g(c) = 0

16. Equivalent statement of IVT • f is continuous on [a,b] • f(a) and f(b) have opposite signs • i.e f(a) ≤ 0 ≤ f(b) or f(b) ≤ 0 ≤ f(a) • then there exists at least one c in [a,b] s.t. f(c) = 0 y y = f(x) f(b) c x a N = 0 b f(a)

17. y 1 x -1 0 1 -1 Continuity is important! • Let f(x) = 1/x • Let a = -1 and b = 1 • f(-1) = -1, f(1) = 1 • However, there is no c such that f(c) = 1/c =0

18. Important remarks • IVT can be used to prove existence of a root of equation • It cannot be used to find exact value of the root!

19. Example 1 • Prove that equation x = 3 – x5 has a solution (root) • Remarks • Do not try to solve the equation! (it is impossible to find exact solution) • Use IVTto prove that solution exists

20. Steps to prove that x = 3 – x5 has a solution • Write equation in the form f(x) = 0 • x5 + x – 3 = 0 so f(x) = x5 + x – 3 • Check that the condition of IVT is satisfied, i.e. that f(x) is continuous • f(x) = x5 + x – 3 is a polynomial, so it is continuous on (-∞, ∞) • Find a and b such that f(a) and f(b) are of opposite signs, i.e. show that f(x) changes sign (hint: try some integers or some numbers at which it is easy to compute f) • Try a=0: f(0) = 05 + 0 – 3 = -3 < 0 • Now we need to find b such that f(b) >0 • Try b=1: f(1) = 15 + 1 – 3 = -1 < 0 does not work • Try b=2: f(2) = 25 + 2 – 3 =31 >0 works! • Use IVT to show that root exists in [a,b] • So a = 0, b = 2, f(0) <0, f(2) >0 and therefore there exists c in [0,2] such that f(c)=0, which means that the equation has a solution

21. x = 3 – x5⇔ x5 + x – 3 = 0 y 31 x 0 2 N = 0 c (root) -3

22. Example 2 • Find approximate solution of the equationx = 3 – x5

23. Idea: method of bisections • Use the IVT to find an interval [a,b] that contains a root • Find the midpoint of an interval that contains root: midpoint = m = (a+b)/2 • Compute the value of the function in the midpoint • If f(a) and f (m) are of opposite signs, switch to [a,m] (since it contains root by the IVT),otherwise switch to [m,b] • Repeat the procedure until the length of interval is sufficiently small

24. f(x) = x5 + x – 3 = 0 We already know that [0,2] contains root f(x)≈ > 0 < 0 31 -3 -1 Midpoint = (0+2)/2 = 1 0 2 x

25. f(x) = x5 + x – 3 = 0 f(x)≈ 31 6.1 -3 -1 1.5 0 2 1 x Midpoint = (1+2)/2 = 1.5

26. f(x) = x5 + x – 3 = 0 f(x)≈ 31 6.1 -3 1.3 -1 0 2 1.5 1 1.25 x Midpoint = (1+1.5)/2 = 1.25

27. f(x)≈ f(x) = x5 + x – 3 = 0 31 6.1 -3 1.3 -.07 -1 1.25 1.125 1 0 2 1.5 Midpoint = (1 + 1.25)/2 = 1.125 x • By the IVT, interval [1.125, 1.25] contains root • Length of the interval: 1.25 – 1.125 = 0.125 = 2 / 16 = = the length of the original interval / 24 • 24 appears since we divided 4 times • Both 1.25 and 1.125 are within 0.125 from the root! • Since f(1.125) ≈ -.07, choose c ≈ 1.125 • Computer gives c ≈ 1.13299617282...

28. Exercise • Prove that the equationsin x = 1 – x2has at least two solutions Hint: Write the equation in the form f(x) = 0 and find three numbers x1, x2, x3, such that f(x1) and f(x2) have opposite signs AND f(x2) and f(x3) haveopposite signs. Then by the IVT the interval [ x1, x2 ] contains a root ANDthe interval [ x2, x3 ] contains a root.