If we recall from the previous section we said that f

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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.

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