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Unformatted text preview: s that we can plug
into a function and get back a real number. At this point, that means that we need to avoid
division by zero and taking square roots of negative numbers.
Let’s do a couple of quick examples of finding domains. Example 4 Determine the domain of each of the following functions.
x+3
(a) g ( x ) = 2
[Solution]
x + 3 x  10
(b) f ( x ) = 5  3 x [Solution]
7x + 8
[Solution]
x2 + 4
10 x  5
(d) R ( x ) = 2
[Solution]
x  16
(c) h ( x ) = Solution
The domains for these functions are all the values of x for which we don’t have division by zero
or the square root of a negative number. If we remember these two ideas finding the domains
will be pretty easy.
(a) g ( x ) = x+3
x + 3x  10
2 So, in this case there are no square roots so we don’t need to worry about the square root of a
negative number. There is however a possibility that we’ll have a division by zero error. To
determine if we will we’ll need to set the denominator equal to zero and solve. x 2 + 3x  10 = ( x + 5 ) ( x  2 ) = 0 x = 5, x = 2 So, we will get division by zero if we plug in x = 5 or...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
 Spring '12
 MrVinh

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