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Determine at exactly which points the following functions are continuous, and at exactly which points the functions are discontinuous. they all have domains[ 0,1]

f(x) =1 if is irrational, and f(x) =0 if x is rational.
g(x) = 0 if z is irrational, and g(x)=x if x is rational.
h(z) = 1-x if x is irrational, and h(x) = x if x is rational.
k(x)= 0 if x-0 or if x is irrational,
while k(m/n)=1/n when x = m/n is a rational number in lowest tenns.

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Final answer:

All the given functions are discontinuous at every point in the interval [0, 1], as they do not meet the criteria for continuity due to jumps between values for rational and irrational numbers.

Step-by-step explanation:

Continuity and Discontinuity of Functions

When analyzing the continuity of the given functions — f(x), g(x), h(x), and k(x) — within their domain [0, 1], we observe the following:

  • The function f(x) is discontinuous at every point in [0, 1], as it takes the value 1 for irrational numbers and 0 for rational numbers, and rational and irrational numbers both densely populate the interval [0, 1].
  • The function g(x) is discontinuous at every point in [0, 1] except at x=0, since for rational x, g(x)=x and for irrational x, g(x)=0, creating a jump at every rational point except at x=0 where the function consistently equals 0 for all types of numbers.
  • The function h(x) is discontinuous at every rational point since for rational x, h(x)=x and for irrational x, h(x)=1-x, creating jumps between these values at rational points. For irrational values, the function is continuous but proves to still be discontinuous when approached by rational sequences.
  • As for k(x), it is discontinuous at every rational point where the value is dictated by the inverse of the denominator in the lowest terms, causing jumps between values for different rational numbers. It equals 0 for irrationals and when x=0.

A function is continuous at a point if the limit as x approaches the point from both sides equals the function's value at that point. The functions given do not satisfy this condition at any point within the interval [0, 1].

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