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
Theorem Suppose that f is continuous at b and 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 called a root of equation
A number c such that f(c)=0 is called a root of 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 c1 c2 c3 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 ≤ ≤ f(b) – N or f(b) – N ≤ ≤ 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) a c x
N = 0 b
f(a)

17. Continuity is important!x y -1 1
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 IVT to 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) = – 3 = -3 < 0
Now we need to find b such that f(b) >0
Try b=1: f(1) = – 3 = -1 < 0 does not work
Try b=2: f(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 2
N = 0
c (root) -3

22. Example 2
Find approximate solution of the equation x = 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 -3 -1 31
Midpoint = (0+2)/2 = 1 2 x

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

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

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

28. Exercise
Prove that the equation sin x = 1 – x2 has 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) have opposite signs. Then by the IVT the interval [ x1, x2 ] contains a root AND the interval [ x2, x3 ] contains a root.