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Continuity (Section 1.8) Alex Karassev Definition A function f is continuous at a number a if lim f ( x) f (a) xa Thus, we can use direct substitution to compute the limit of function that is continuous at a Some remarks Definition of continuity requires three things: f(a) is defined (i.e. a is in the domain of f) lim f ( x ) exists x a Limit is equal to the value of the function The graph of a continuous functions does not have any "gaps" or "jumps" Continuous functions and limits Theorem Suppose that f is continuous at b g ( x) b and lim Then xa lim f ( g ( x)) f (lim g ( x)) x a x a Example lim x 2 x2 4 x2 4 ( x 2)( x 2) lim lim x 2 x 2 x 2 x2 x2 lim ( x 2) 2 2 4 2 x 2 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: 1 f ( x) 2 x 1 cos 2 x Solution 1 f ( x) 2 x 1 cos 2 x 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 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 a cb 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) a c1 c2 c3 x 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 N=0 c x b f(a) Continuity is important! y 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 1 x -1 0 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 Check that the condition of IVT is satisfied, i.e. that f(x) is continuous x5 + x – 3 = 0 so f(x) = x5 + x – 3 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 N=0 0 -3 2 x c (root) 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)≈ -3 0 x <0 -1 Midpoint = (0+2)/2 = 1 >0 31 2 f(x) = x5 + x – 3 = 0 f(x)≈ -3 0 x -1 6.1 31 1 1.5 2 Midpoint = (1+2)/2 = 1.5 f(x) = x5 + x – 3 = 0 f(x)≈ -3 0 x -1 1.3 6.1 31 1 1.25 1.5 2 Midpoint = (1+1.5)/2 = 1.25 f(x) = x5 + x – 3 = 0 f(x)≈ -3 -1 -.07 1.3 1 1.125 0 x 1.25 6.1 31 1.5 2 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.