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28 5. CONTINUITY
5.2. Exercises
(1) Let f(x) =
x3 − 2x2 − 2x− 3
x3 − 4x2 + 4x− 3
for x 6= 3. How should f be defined at x = 3 so that it
becomes a continuous function on all of R?
Answer: f(3) =
a
7
where a = .
(2) Let f(x) =

1 if x 3
.
(a) Is it possible to define f at x = 0 in such a way that f becomes continuous at x = 0?
Answer: . If so, then we should set f(0) = .
(b) Is it possible to define f at x = 1 in such a way that f becomes continuous at x = 1?
Answer: . If so, then we should set f(1) = .
(c) Is it possible to define f at x = 3 in such a way that f becomes continuous at x = 3?
Answer: . If so, then we should set f(3) = .
(3) Let f(x) =

x+ 4 if x 3
.
(a) Is it possible to define f at x = −2 in such a way that f becomes continuous at
x = −2? Answer: . If so, then we should set f(−2) = .
(b) Is it possible to define f at x = 1 in such a way that f becomes continuous at x = 1?
Answer: . If so, then we should set f(1) = .
(c) Is it possible to define f at x = 3 in such a way that f becomes continuous at x = 3?
Answer: . If so, then we should set f(3) = .
(4) The equation x5 + x3 + 2x = 2x4 + 3x2 + 4 has a solution in the open interval (n, n+ 1)
where n is the positive integer .
(5) The equation x4−6x2−53 = 22x−2x3 has a solution in the open interval (n, n+1) where
n is the positive integer .
(6) The equation x4 + x + 1 = 3x3 + x2 has solutions in the open intervals (m,m + 1) and
(n, n+ 1) where m and n are the distinct positive integers and .
(7) The equation x5 + 8x = 2x4 + 6x2 has solutions in the open intervals (m,m + 1) and
(n, n+ 1) where m and n are the distinct positive integers and .