While we are on the subject of function evaluation we

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Unformatted text preview: [Solution] ( ) [Solution] (g) g ( x - 5 ) [Solution] (f) f x 3 2 © 2007 Paul Dawkins 178 http://tutorial.math.lamar.edu/terms.aspx College Algebra Solution (a) f ( 3) and g ( 3) Okay we’ve got two function evaluations to do here and we’ve also got two functions so we’re going to need to decide which function to use for the evaluations. The key here is to notice the letter that is in front of the parenthesis. For f ( 3) we will use the function f ( x ) and for g ( 3) we will use g ( x ) . In other words, we just need to make sure that the variables match up. Here are the evaluations for this part. f ( 3) = ( 3) - 2 ( 3) + 8 = 9 - 6 + 8 = 11 2 g ( 3) = 3 + 6 = 9 = 3 [Return to Problems] (b) f ( -10 ) and g ( -10 ) This one is pretty much the same as the previous part with one exception that we’ll touch on when we reach that point. Here are the evaluations. f ( -10 ) = ( -10 ) - 2 ( -10 ) + 8 = 100 + 20 + 8 = 128 2 Make sure that you deal with the negative signs properly here. Now the second one. g ( -10 ) = -10 + 6 = -4 We’ve now reached the difference. Recall that when we first started talking about the definition of functions we stated that we were only going to deal with real n...
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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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