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CONTINUITY In an informal way, we can say that a function f is continuous on an interval if its graph can be drawn without taking the pencil off of the.

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Presentation on theme: "CONTINUITY In an informal way, we can say that a function f is continuous on an interval if its graph can be drawn without taking the pencil off of the."— Presentation transcript:

1 CONTINUITY In an informal way, we can say that a function f is continuous on an interval if its graph can be drawn without taking the pencil off of the paper ( f doesn’t cut or jump). More precisely f is continuous at x = c if and only if: I) f(c) is defined, and II) lim f( x) = f( c ). x x c (meaning you can get to the point (c,f(c) ) by walking on the curve from left or from the right of it) A function f is said to be continuous on an interval I (domain) if and only if f is continuous at any point in I.I. Ex 1. Find where f(x) = (2x-4) / (x 2 –x –6) is not continuous ( points of discontinuity ). Since f(x) is not defined at x where its denominator is zero. The discontinuity of f occurs when x 2 –x –6 = 0. Solving x 2 –x –6 = 0, we get x=-2 or x= 3. So the values for which f is discontinuous are x=-2 or x=3. Ex 2. The line f(x) = ax+b is continuous in its domain R. f(c) = ac+b exists and lim.. x x c (ax+b) =ac+b = f( c ), so f is continuous at any real c Ex 3. Polynomials p(x) are continuous on R, since lim p(x) = p(a) and p(x)| x=a = p(a). xaxa Note that if f is continuous on I, then lim f(x) coincides with f(c) x  c

2 y= f(x) C0NTINUOUS FUNTIONS V/S DISCONTINUOUS FUNCTIONS Analyze continuity for each of the four functions illustrated with the graphs shown below. 1) y = f(x) is continuous, since f has no disconnecting places or jumps. 2) y = g(x) is continuous, since g has no disconnecting places or jumps. 3) y = h(x) is discontinuous, since h has a jump at x = a. 4) y = r(x) is discontinuous, since r is disconnected at x=c. y= g(x) Jump y=h(x) There’s a hole. A point is missing! Y X c y=r(x)

3 y = x y = 3x MORE PROPERTIES OF CONTINUOUS FUNCTIONS Theorem: (Combination of continuous functions). If f and g are continuous at x=c then the following combinations of them are continuous at x=c too. 4) h(x) = (x 2 - 3x + 4 ) / (x-2) Example 1: We know f(x) = x is continuous. 2) ( f  f)(x) = f(x)  f(x) = x  x = x2x2 3) g((x) =( f f)(x)+4 f(x)+5 = x 2 - 4x y = x 2 ` 5) Composition: f o g is continuous. (provides that f is continuous at g(c) ) 4) Quotient: f/g is continuous if g(c) ≠ 0 3) Product: f  g is continuous 2) Sum and Difference: f +g and f - g are continuous 1) Scalar multiple: kf is continuous, for any constant k The following functions are continuous too. See figure on the right. 1) 3f(x) = 3x 2 y=x 2 - 4x+ 5 X Y 2 Note that x =2 is a Vertical Asymptote for h(x) So h(x) is continuous for any x  2. And the oblique asymptote is y=x-1. y=x-1

4 The above examples show us that polynomial functions are continuous functions on R and that the rational functions q(x) = p 1 (x) / p 2 (x) (where p 1 & p 2 are polynomials) are continuous on R – { the zeros of p 2 (x) }. Example 1: Study continuity for f(x) = (x+1)/ (x 2 – x –12) We look for the zeros of denominator, i.e. x 2 –x–12 =0  (x +3)(x – 4)=0. So x=4 or x= - 3. Example 2: Study continuity for g(x) = sin 2 x /(x 4 + 1). (Assume that sinx is continuous) Since sin x is continuous its, square is also continuous. So f(x) is continuous on R – {-3,4} Now not only that the denominator is a continuous function (it’s a polynomial ! ). Furthermore x cannot be zero, since x is always  1 so. So g(x) is continuous on R. Example 3: Study continuity for sin x & cos x But we know that lim sin = 0 & lim cos h = 1. h  0 h  0 lim sin(x+h) = lim (sin x cos h + cos x sinh). h  0. h  0 Since sin(x+h) = sin x cos h + cos x sinh we have = sin x lim cos h + cos x lim sin h.. h  0 h  So lim sin(x+h) = sin x and this means that sin x is a continuous function.. h  0 Similarly cos x is a continuous function using cos(x+h) = cos x cos h - sinx sinh

5 Theorem: If f(x) is a continuous function on a closed interval [a,b], then f(x) must take all values between f(a) and f(b). f(a) f(b) a b c The figure on the right shows a continuous function on [a,b] The Theorem assures us that if we choose any number c between f(a) and f(b), the line y=c will have to intersect at least one point on the graph. In our case it cuts the graph at three points (x 1, c), (x 2, c) & (x 3, c). x 1 x 2 x 3 X Y Intermediate Value Theorem Example: Prove that f(x) = x 3 +5x – 7 has a zero between 1 and 2. Since f(1) = = - 5 & f(2) = = 7. We have f(1)<0

6 2 Theorem: If f is continuous at g(x) then lim f( g(x) ) = f( lim (g(x) )... x  a x  a Interchanging lim and ln, we get ln [ lim ((3x+1)/(x+9))]. x x  2 Example 2: Find lim. x   = ln(3) ln [(3x+1)/(x+9)] (Assume that ln x is continuous for x>0) Example 1: Find lim f(x),. x  16 for f(x) = x 3/2. (Assume that f(x) is continuous) Interchanging lim and f we get lim f(x) = [ lim x ] 3/2 =16 3/2 = 64 x  16 x  16 Example 3: Find lim f(x),. x  -1 for f(x) = (sin [  (x 2 – 1)/(x 2 +4x+3) ] ) 2 Interchanging lim and ( ) 2 …. {lim sin[  (x 2 – 1)/(x 2 +4x+3) } 2.. x  -1 Interchanging lim and sin ….. {sin lim  (x 2 – 1)/(x 2 +4x+3) } 2.. x  -1 But lim  (x 2 – 1)/(x 2 +4x+3) is type 0/0, so we try to simplify by (x+1) ….. x  -1 Factoring we get lim  (x 2 – 1)/(x 2 +4x+3) = lim  (x+1)(x-1) / (x+1)(x+3). x  -1 x  -1 = -  So lim f(x) = (sin (-  )) 2 = 0. x  -1


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