# 3.2 -     MATH 1081  Monday, January 31   ...

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Unformatted text preview:     MATH 1081  Monday, January 31    Chapter 3 – Section 2  CONTINUITY  Homework #3 (due 2/7):    Section 3.2 #6, 20, 26                  Section 3.3 #8, 30, 34    Clicker Check­In:  Choose any letter to check in now. Now, we saw in section 3.1 that for many (although certainly  not all) functions at most choices of a,                   lim f ( x ) = f ( a).        That is, the value of the limit coincides with the value of the  function.  So, the point at  x = a is where we would “expect” it to  be.  x →a For us, this is precisely the criterion for continuity of a function  at a point.      A very informal notion of continuity at a point is that there is  no hole or gap or jump or asymptote.  The curve can be drawn  through that point without lifting the pencil.      This is what this limit equation also says…as  x  gets close to  a ,  the values of the function are getting close to a limiting value  AND that happens to be exactly equal to the value that the  function takes on at  x = a.  The book describes this with a 3‐part definition.    A function f is continuous at  x = a if the following 3 conditions  are all satisfied:    1.   f ( a)  is defined.    (So the function has a value at  x = a.)    2.   lim f ( x )  exists.    (So the function is tending to one specific value.)    3.   lim f ( x ) = f ( a).    x →a x →a (The value of the function and the limit are the same.)  Example:  Determine if the function is continuous at the   indicated point.  If not, state which of the     criteria fail.    1. f ( x ) = 9 x 2 + 5   at   x = 1    x2 − 4 2. f ( x ) =    x−2 at   x = 2  at   x = 3    Note, if a function is not continuous at a point, it is said to be  discontinuous there. x+9 3. f ( x ) =    x−3     • Polynomial functions are continuous for all  x .  • Rational functions are continuous for all  x  in the domain.  • Exponential functions are continuous for all  x .  • Logarithmic functions are continuous for all  x  in the  domain.  • Radical (or Root) functions are continuous for all  x  in the  domain.  Now, as we observed last time,  lim x  doesn’t exist.  So, can  the function  f ( x ) = x  continuous at  x = 0 ?        In the sense that there is no hole, gap, jump, vertical asymptote  at  x = 0  it satisfies the intuitive notion of continuous.    However, in the definition that  lim f ( x ) = f ( 0 ), we have an  issue, since the limit does not exist.    x→ 0 x→ 0 So, to deal with this and other like situations, we can define a  function as continuous on a closed interval [ a, b ] if    1. It is continuous at all points in the open interval ( a, b ).  2. It is continuous from the right at  x = a .  That is,  lim+ f ( x ) = f ( a ) .  3. It is continuous from the left at  x = b .  That is,  lim− f ( x ) = f ( b ) .    Extending this in the obvious way, we can say that  f ( x ) = x  is  continuous on its domain [ 0, ∞ ).    x→b x→ a A particular kind of function that requires a bit more analysis is  one that is defined not as a single algebraic expression, but  maybe two or more pieces.  These are referred to as  (appropriately enough) piecewise­defined functions.    For example,    is such a function.  To the left side of  x = 2 , the function  behaves like the function  f1 ( x ) = x + 4 .  To the right side of  x = 2 , the function behaves like  f2 ( x ) = x 2 + 1.  ⎧ x + 4 for f ( x) = ⎨ 2 ⎩ x + 1 for x<2   x≥2 Both  f1 ( x ) = x + 4  and  f2 ( x ) = x 2 + 1 are polynomials which are  continuous everywhere.  So the only real questionable point of  ⎧ x + 4 for continuity for  f ( x ) = ⎨ 2 ⎩ x + 1 for x→ 2 x<2  is at  x = 2 .  x≥2   To evaluate  lim f ( x ) , we MUST analyze each one‐sided limit  separately, since the function is defined by a different  expression to the left of  x = 2  than to the right of  x = 2 .    What are these one‐sided limits?    Is  f ( x )  continuous at  x = 2 ?  Example:  Determine all values of  x  where the function is         discontinuous.  Which of the 3 criteria for    continuity fails in each case?    ⎧2 if x < −1 ⎪       1.   f ( x ) = ⎨ 3 + x if −1 < x ≤ 3   ⎪ 4 − x if x>3 ⎩   ⎧ x 2 + 1 if x < 0 ⎪       2.   g ( x ) = ⎨ −1 if x = 0   ⎪ ex if x > 0 ⎩   Clicker Check­out: Choose any letter to check out now.    Next time – Section 3.3     ...
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## This note was uploaded on 05/01/2011 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.

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