A function f(x) is said to have a removable discontinuity at x = a if: 1. f is either not defined or not continuous at x = a. 2. f(a) could either be defined or redefined so that the new function IS continuous at x = a. Let 4 - 3x + 20 if x + 0 and x + 5 x ( x – 5) f(x) = 4 | if x = 0 Show that f(x) has a removable discontinuity at 0 and determine what value for f(0) would make f(x) continuous at x = 0. Must redefine f(0) Hint: Try combining the fractions and simplifying. The discontinuity at x = removable discontinuity, just in case you were wondering. 5 is actually NOT a

College Algebra (MindTap Course List)
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ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...
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A function f(x) is said to have a removable
discontinuity at x = a
1. f is either not defined or not continuous at x = a.
2. f(a) could either be defined or redefined so that
the new function IS continuous at x = a.
Let
- 3x + 20
+
x (x – 5)
4
if x + 0 and x + 5
f(x)
4
if x =
=D0
Show that f(x) has a removable discontinuity at
0 and determine what value for f(0) would
make f(x) continuous at x =
0.
Must redefine f(0)
Hint: Try combining the fractions and simplifying.
The discontinuity at x =
removable discontinuity, just in case you were
wondering.
5 is actually NOT a
Transcribed Image Text:A function f(x) is said to have a removable discontinuity at x = a 1. f is either not defined or not continuous at x = a. 2. f(a) could either be defined or redefined so that the new function IS continuous at x = a. Let - 3x + 20 + x (x – 5) 4 if x + 0 and x + 5 f(x) 4 if x = =D0 Show that f(x) has a removable discontinuity at 0 and determine what value for f(0) would make f(x) continuous at x = 0. Must redefine f(0) Hint: Try combining the fractions and simplifying. The discontinuity at x = removable discontinuity, just in case you were wondering. 5 is actually NOT a
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