Download presentation

Published byKiera Dimit Modified over 3 years ago

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 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. Note that if f is continuous on I, then lim f(x) coincides with f(c) x c Ex 1. Find where f(x) = (2x-4) / (x2 –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 x2 –x –6 = 0. Solving x2 –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 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) xa

2
**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. Y y=h(x) y= f(x) 2) y = g(x) is continuous, since g has no disconnecting places or jumps. y= g(x) y=r(x) 3) y = h(x) is discontinuous , since h has a jump at x = a. Jump X 4) y = r(x) is discontinuous, since r is disconnected at x=c. c There’s a hole. A point is missing!

3
**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. 1) Scalar multiple: kf is continuous, for any constant k 2) Sum and Difference: f +g and f - g are continuous 3) Product: fg is continuous 4) Quotient: f/g is continuous if g(c) ≠ 0 y = 3x 5) Composition: fog is continuous (provides that f is continuous at g(c) ) y = x2 ` Y Example 1: We know f(x) = x is continuous. The following functions are continuous too See figure on the right. y=x2 - 4x+ 5 y = x y=x-1 1) 3f(x) = 3x 2) ( ff)(x) = f(x)f(x) = xx = x2 X 3) g((x) =( f•f)(x)+4 f(x)+5 = x2 - 4x + 5 2 2 2 4) h(x) = (x2 - 3x + 4 ) / (x-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.

4
The above examples show us that polynomial functions are continuous functions on R and that the rational functions q(x) = p1(x) / p2(x) (where p1 & p2 are polynomials) are continuous on R – { the zeros of p2(x) }. Example 1: Study continuity for f(x) = (x+1)/ (x2 – x –12) We look for the zeros of denominator, i.e. x2–x–12 =0 (x +3)(x – 4)=0. So x=4 or x= - 3. So f(x) is continuous on R – {-3,4} Example 2: Study continuity for g(x) = sin2 x /(x4 + 1). (Assume that sinx is continuous) Since sin x is continuous its, square is also continuous. Now not only that the denominator is a continuous function (it’s a polynomial ! ). Furthermore x cannot be zero, since x4 + 1 is always 1 so . So g(x) is continuous on R. Example 3: Study continuity for sin x & cos x Since sin(x+h) = sin x cos h + cos x sinh we have 1 lim sin(x+h) = lim (sin x cos h + cos x sinh) h h0 = sin x lim cos h + cos x lim sin h h h0 But we know that lim sin = 0 & lim cos h = h h0 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
**Intermediate Value Theorem**

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). Y f(a) f(b) The figure on the right shows a continuous function on [a,b] c 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 (x1, c) , (x2 , c) & (x3 , c). X a b x x2 x3 Example: Prove that f(x) = x3+5x – 7 has a zero between 1 and 2. Y (2,7) 7 We know that f(x) = x3+5x – 7 is a continuous function. X Since f(1) = = - 5 & f(2) = = 7. We have f(1)<0<f(2). - 5 (1,-5) The Intermediate Theorem assures us that the line y=0 intersects the graph of f(x) at a value of x between 1 and 2.

6
**Example 1: Find lim f(x) , . x 16 **

Theorem: If f is continuous at g(x) then lim f( g(x) ) = f( lim (g(x) ) xa xa Example 1: Find lim f(x) , x 16 for f(x) = x3/2 . (Assume that f(x) is continuous) Interchanging lim and f we get lim f(x) = [ lim x ]3/2 =163/2 = x x 16 Example 2: Find lim x ln [(3x+1)/(x+9)] (Assume that ln x is continuous for x>0) Interchanging lim and ln, we get ln [ lim ((3x+1)/(x+9))] x = ln(3) Example 3: Find lim f(x) , x-1 for f(x) = (sin [(x2 – 1)/(x2+4x+3) ] )2 Interchanging lim and ( )2 …. {lim sin[(x2 – 1)/(x2+4x+3) } x -1 Interchanging lim and sin ….. {sin lim (x2 – 1)/(x2+4x+3) } x-1 But lim (x2 – 1)/(x2+4x+3) is type 0/0 , so we try to simplify by (x+1) … x-1 2 2 Factoring we get lim (x2 – 1)/(x2+4x+3) = lim (x+1)(x-1) / (x+1)(x+3) x x -1 = - So lim f(x) = (sin (-))2 = x-1

Similar presentations

OK

LIMITS 2. We noticed in Section 2.3 that the limit of a function as x approaches a can often be found simply by calculating the value of the function.

LIMITS 2. We noticed in Section 2.3 that the limit of a function as x approaches a can often be found simply by calculating the value of the function.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on unified power quality conditioner Ppt on electricity for class 10 download Ppt on event handling in javascript Ppt on stem cell technology Cell surface display ppt online Ppt on fire extinguisher types video Free ppt on personality development Ppt on articles of association hong Ppt on service oriented architecture Ppt on history and sport the story of cricket