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Continuity of Composite Functions

F(x) = f(g(x)) is discontinuous also at those values of x where g(x) is discontinuous. For example, f(x) = 1/(1 – x) is discontinuous at x = 1. Now,
f(f(x)) = 87972.png
 
is not only discontinuous at x = 0 but also at x = 1. Now,
f(f(f(x))) = 87966.png
 
seems to be continuous, but it is discontinuous at x = 1 and x = 0 where f(x) and f(f(x)) respectively are discontinuous.
 
Illustration
If f(x) = 87960.png and g(x) = 87954.png, then discuss the continuity of f(x), g(x), and fog(x).
Solution
f(x) = 87947.png
 
f is not defined at x = 1, therefore, f is discontinuous at x = 1.
 
g(x) = 87941.png
 
g(x) is not defined at x = 2, therefore, g is discontinuous at x = 2.
 
Now fog(x) will be discontinuous at
  1. x = 2 (point of discontinuity of g(x)).
  2. g(x) = 1 (when g(x) = point of discontinuity of f(x)).
If g(x) = 1 ⇒ 1/(x – 2) = 1 ⇒ x = 3, therefore, fog(x) is discontinuous at x = 2 and x = 3. Also
 
fog(x) = 87928.png
 
and fog(2) is not defined.
87922.png = 87916.png
87910.png
 
Therefore, fog(x) is discontinuous at x = 2 and it has removable discontinuity at x = 2 .
 
For continuity at x = 3,
87904.png = 87898.png
87892.png = 87886.png
 
Therefore, fog(x) is discontinuous at x = 3 and it has non-removable discontinuity at x = 3,
 

Properties of functions continuous in [a, b]

  1. If a function f is continuous on a closed interval [a, b] then it is bounded.
  2. A continuous function whose domain is some closed interval must have its range also in closed interval.
  3. If f(a) and f(b) possess opposite signs then at least one solution of the equation f(x) = 0 in the open interval (a, b) provided f is continuous in [a, b].
  4. If f is continuous on [a, b] then f–1 (from the range of f ) is also continuous.




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