- 51 Views
- Uploaded on
- Presentation posted in: General

Continuity ( Section 1.8)

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Continuity (Section 1.8)

Alex Karassev

- 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

- 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"

- TheoremSuppose that f is continuous at bandThen
- Example

- 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.

- 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

- Determine, where is the following function continuous:

- 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)

Intermediate Value Theorem

- A solution of equation is also calledarootof equation
- A number c such that f(c)=0 is calledarootof function f

- 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

- 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

- 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

- 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)

y

1

x

-1

0

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

- IVT can be used to prove existence of a root of equation
- It cannot be used to find exact value of the root!

- 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

- 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

y

31

x

0

2

N = 0

c (root)

-3

- Find approximate solution of the equationx = 3 – x5

- 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

We already know that [0,2] contains root

f(x)≈

> 0

< 0

31

-3

-1

Midpoint = (0+2)/2 = 1

0

2

x

f(x)≈

31

6.1

-3

-1

1.5

0

2

1

x

Midpoint = (1+2)/2 = 1.5

f(x)≈

31

6.1

-3

1.3

-1

0

2

1.5

1

1.25

x

Midpoint = (1+1.5)/2 = 1.25

f(x)≈

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...

- 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.