sketch the graph of t. does t have any discontinuities? explain why, if t had a jump discontinuity, it might be advantageous in some situations to earn less money.
further insights and challenges
94. if f has a removable discontinuity at x = c, then it is possible to redefine f(c) so that f is continuous at x = c. can this be done in more than one way? explain.
95. give an example of functions f and g such that f(g(x)) is continuous but g has at least one discontinuity.
96. continuous at only one point show that the following function is continuous only at x = 0:
f(x) = {x for x rational; -x for x irrational}
97. show that f is a discontinuous function for all x, where f(x) is defined as follows:
f(x) = {1 for x rational; -1 for x irrational}
show that f² is continuous for all x.

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# Brief Explanations for 94: A removable discontinuity at $ x = c $ implies $ \lim_{x \to c} f(x) $ exists (unique $ L $). To make $ f $ continuous at $ c $, $ f(c) $ must equal $ L $, so only one way to redefine $ f(c) $.
# Answer for 94: No, it cannot be done in more than one way. A removable discontinuity at $ x = c $ requires $ \lim_{x \to c} f(x) = L $ (unique), so $ f(c) $ must equal $ L $ for continuity, leaving only one possible value for $ f(c) $.

# Brief Explanations for 95: Choose $ g $ with a discontinuity and $ f $ continuous such that $ f(g(x)) $ is constant. Example: $ g(x) $ has a jump discontinuity, $ f(x) = x^2 $.
# Answer for 95: Let $ g(x) = \begin{cases} 1 & x \geq 0 \ -1 & x < 0 \end{cases} $ (discontinuous at $ x = 0 $) and $ f(x) = x^2 $ (continuous). Then $ f(g(x)) = (1)^2 = 1 $ for $ x \geq 0 $ and $ (-1)^2 = 1 $ for $ x < 0 $, so $ f(g(x)) = 1 $ (continuous everywhere).

# Brief Explanations for 96: At $ x = 0 $, $ \lim_{x \to 0} f(x) = 0 = f(0) $. For $ x
eq 0 $, rational/irrational sequences have conflicting limits.
# Answer for 96: For $ x = 0 $: $ \lim_{x \to 0} f(x) = 0 $ (since $ |f(x)| = |x| \to 0 $) and $ f(0) = 0 $, so continuous. For $ x
eq 0 $: Take rational $ r_n \to x $ (lim $ f(r_n) = x $) and irrational $ s_n \to x $ (lim $ f(s_n) = -x $); $ x
eq -x $, so no limit, discontinuous. Thus, continuous only at $ x = 0 $.

# Brief Explanations for 97: $ f $ has conflicting limits (1/-1) at any $ x $; $ f^2(x) = 1 $ (constant, continuous).
# Answer for 97: $ f $ is discontinuous for all $ x $: For any $ x $, rational $ r_n \to x $ (lim $ f(r_n) = 1 $) and irrational $ s_n \to x $ (lim $ f(s_n) = -1 $); 1 ≠ -1, so no limit. $ f^2(x) = 1 $ for all $ x $ (constant function), which is continuous everywhere.

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