Intermediate Value Theorem

Alex Karassev

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.