Intermediate Value Theorem Alex Karassev

River and Road

River and Road

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

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

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

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 y y = f(x) f(b) a c x N = 0 b f(a)

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

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 y 31 x 2 N = 0 c (root) -3

Example 2 Find approximate solution of the equation x = 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 We already know that [0,2] contains root f(x)≈ < 0 > 0 -3 -1 31 Midpoint = (0+2)/2 = 1 2 x

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

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

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 = (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) have opposite signs. Then by the IVT the interval [ x1, x2 ] contains a root AND the interval [ x2, x3 ] contains a root.