3/11/10

Test 5 Section 4 - #13 (page 654)

The first thing we want to do any time we have weird stuff like 2 f(p) = 20 is get the f(p) out of there:

f(x) = x+1, 
so f(p)=p+1

Now let's work on solving for p:

2(p+1)=20
p+1 = 10
p=9

We're not done yet, of course.  The question asked us for f(3p), which we can now rewrite as f(27).  Just plug 27 into the function like we did with p before:

f(27) = 27 + 1
f(27) = 28

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