Continuity (Section 1.8)
Alex Karassev
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
)()(lim afxfax
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"
)(lim xfax
Continuous functions and limits
TheoremSuppose that f is continuous at band Then
Example
bxgax
)(lim
))(lim())((lim xgfxgfaxax
2422)2(lim
2
)2)(2(lim
2
4lim
2
4lim
2
2
2
2
2
2
x
x
xx
x
x
x
x
x
xxx
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.
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
Example
Determine, where is the following function continuous:
x
xxf2
1cos12)(
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)
x
xxf2
1cos12)(
Intermediate Value Theorem
River and Road
River and Road
Definitions
A solution of equation is also calleda root of equation
A number c such that f(c)=0 is calleda root of function f
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
x
y
a
y = f(x)
f(a)
f(b)
b
N
c
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
x
y
a
y = f(x)
f(a)
f(b)
b
N
c1 c2c3
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
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
x
y
a
y = f(x)
f(a)
f(b)
bN = 0
c
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
x
y
0-1
-1
1
1
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!
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
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
x = 3 – x5 ⇔ x5 + x – 3 = 0
x
y
0
-3
31
2N = 0
c (root)
Example 2
Find approximate solution of the equationx = 3 – x5
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
f(x) = x5 + x – 3 = 0
0 2
f(x)≈
x
-3 31
We already know that [0,2] contains root
Midpoint = (0+2)/2 = 1
-1< 0 > 0
f(x) = x5 + x – 3 = 0
0 2
f(x)≈
x
-3 31
1
-1
1.5
6.1
Midpoint = (1+2)/2 = 1.5
f(x) = x5 + x – 3 = 0
0 2
f(x)≈
x
-3 31
1
-1
1.5
6.1
Midpoint = (1+1.5)/2 = 1.25
1.25
1.3
1
-1
1.25
1.3
1.125
-.07
f(x) = x5 + x – 3 = 0
0 2
f(x)≈
x
-3 31
1.5
6.1
Midpoint = (1 + 1.25)/2 = 1.125
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...
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) haveopposite signs. Then by the IVT the interval [ x1, x2 ] contains a root ANDthe interval [ x2, x3 ] contains a root.
