f(x+1) = f(x) + 1
f(7) = f (6+1) = f(6) + 1 = 3 ---> f(6) = 2
f(6) = f (5+1) = f(5) + 1 = 2 ---> f(5) = 1
f(5) = f (4+1) = f(4) + 1 = 2 ---> f(4) = 0
f(4) = f (3+1) = f(3) + 1 = 2 ---> f(3) = -1
Option A is the answer.
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RedeemIf \(f(x+1)=f(x)+1\) for all values of \(x,\) and \(f(7)=3,\) what is the value of \(f(3)?\)
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Solution:
Letting x = 6, we have:
f(6 + 1) = f(6) + 1
f(7) = f(6) + 1
3 = f(6) + 1
2 = f(6)
Now letting x = 5, we have:
f(5 + 1) = f(5) + 1
f(6) = f(5) + 1
2 = f(5) + 1
1 = f(5)
Using the same argument, we see that f(4) = 0 and f(3) = -1.
Answer: A
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