1. Intermediate Value Theorem
Alex Karassev

2. River and Road

3. River and Road

4. 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

5. 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

6. 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

7. 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

8. 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)

9. 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

10. 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!

11. 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

12. 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

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

14. Example 2
Find approximate solution of the equation x = 3 – x5

15. 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

16. 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

17. 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

18. 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

19. 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 ≈

20. 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.