Unformatted text preview: , let’s get a little more complicated, or at least they appear to be more complicated. Things
aren’t as bad as they may appear however. We’ll evaluate f ( t + 1) first. This one works exactly
the same as the previous part did. All the x’s on the left will get replaced with t + 1 . We will
have some simplification to do as well after the substitution. f ( t + 1) = ( t + 1) - 2 ( t + 1) + 8
2 = t 2 + 2t + 1 - 2t - 2 + 8
= t2 + 7
Be careful with parenthesis in these kinds of evaluations. It is easy to mess up with them.
Now, let’s take a look at f ( x + 1) . With the exception of the x this is identical to f ( t + 1) and
so it works exactly the same way. f ( x + 1) = ( x + 1) - 2 ( x + 1) + 8
2 = x2 + 2 x + 1 - 2 x - 2 + 8
= x2 + 7
Do not get excited about the fact that we reused x’s in the evaluation here. In many places where
we will be doing this in later sections there will be x’s here and so you will need to get used to
[Return to Problems] () (f) f x 3 Again, don’t get excited about the x’s in the parenthesis here. Just evaluate it as if it were a
number. f ( x3 ) = ( x 3 ) - 2 ( x 3 ) + 8
2 = x 6 - 2 x3 + 8
[Return to Problems] ( (g) g x 2 - 5 ) One more evaluation and this time we’ll use the o...
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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