2.3 - Continuity

2.3 - Continuity - Limits are technicaly defined independantly of the value of a function. 1. e. lim f(x) has NOTHING to do with f(a) xa uniess the function f(x) is CONTINUOUS (i.c. polynomials) Definition f(a) = y coordinate If① lim f(x) exists and ② f(a) is defined and lim f(x) = f(a) then f(x) is continuous at xa Note: The flip-side of this definition teus us that: If any of those 3 conditions IS NOT SATISFIED then f(x) IS NOT CONTINUOUS (DISCONTINUOus) * If there is a vertical asymptote and your "mit goes to ± 0∞, the limit doesn't actually exist (in this case) because it is not a real number. (you will have to take your pen off the paper to araw the rest of the function ex : if limf(x) does not exist X-> a if f(a) is undefined if (3mf(x) - +(a) The following are wwh to be continuous for all real numbers x I constant functions : polynomial functions (which include linear s quadratic 3 rational functions EXCEPT at values of x that make denominator 0 . 4 root functions : f(x) =YEXCEPT at values of x that make the inside NEGATVE when n is EVEN exponentials logarithmic functions EXCEPT at values of x that make a logarithm undefined ⑥ All sums/differences/products/Quotients of continuous are also continuous ! A function f(x) is continuous at x a if ... 1 lim f(x) exists 2 f(a) is defined Slim f(x) and f(a) are equal A function is Not continuous & x = a if ... ·imf(x) DNE x-a 2 f(a) is not defined 3 limf(x) and f(a) are not equal xa ex: Let 2 2 4 This function is not continuous at: 2 limf(x) = f(2) X = - 2 limf(x) DNE x = 5 lim f(x) DNE and fl-5) is not defined b limf(x) is not defined The function Is continuous over the intervals : Recall : polynomials , rationals , roots , exponentials , logarithms are ALL CONTINUOUS at every value x = a in their domains So are all sums , differences, products + quotients of these functions ex : Find all values of X where the following functions are ot continuous . This is a RATIONAL function 24 it is continuous at every x except those that are Not in its domain i .e. values of X that make 4 The function is Not continuous -1 NO SOLUTION : This function Is continuous for all real numbers (3) h(x) = { 1+2x 5-2x if x < 2 if x 2 This is a piecewise function, made up of LINEAR functions (polynomials) 1+2x 5-2x is continuous for x 2 continuous for x 22. but what happens e x = 2 ??? does lim (n(x)) exist? X2 lim n(x) x2 lim (1+2) 12(2) S lim h(x) D.NE 2 lim n(x) = lim (5-2x) X+2+ = 5-2(2) h(x) is NOT continuous at x=2 FOCUSING ON FUNCTIONS THAT ARE CONTINUOUS Question: how do you calculate the slope of a ivne? m = Уг У. = ΔΥ how do you calculate the slope at a point on a curve? on a line, the slope is constant on a curve, the slope is always changing but we can "approximate" the slope at a point on a curve by using a second point close to the first. Given a curve y=f(x) If you want to know the SLOPE at the point P(a, fla)) let Q be a second point with coordinates Q(ath, f(ah)) the slope of the line through P. a 15: M = AY f(a+h) fla) ath- M = f(ath) f(a) this is called the slope of a secant line we can now let heo using LIMITS, and the result will be the slope of the TANGENT LINE at the point p. lim The result is called THE DERIVATIVE of f(x) asxa ex: Ict f(x) = x² (a) calculate the slope of the secant line through the points (1,1(1)) and (1.5,f(is)) = (1.5)-(1)² = 0.625 (b) same as (a) for (1, 1C1)) $ (1+n, t(in)) m= f(i+n)-f(1) 1th-1 = (1+n)²-4(1)² = 4 (1.2n+n²)-4 = ++²